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DIFFERENTIAL    AND    INTEGRAL    CALCULUS 


THE  MACIMILLAN  COMPANY 

NEW  YORK    •    BOSTON   ■    CHICAGO   •    DALLAS 
ATLANTA  •    SAN   FRANCISCO 

MACMILLAN  &  CO.,  Limited 

LONDON  •  BOMBAY  •  CALCUTTA 
MELBOURNE 

THE  MACMILLAN  CO.  OF  CANADA,  Ltd. 

TORONTO 


DIFFERENTIAL  AND  INTEGRAL 


CALCULUS 


BY 


CLYDE    E.    LOVE,    Ph.D. 

ASSISTANT    PROFESSOR    OF    MATHEMATICS    IN    THE 
UNIVERSITY    OF    MICHIGAN 


'j     ^„o     J     i      i> 


l^efe  gnrit 

THE   MACMILLAN    COMPANY 

1917 

All  rights  reserved 


^ 


hlol 


Copyright,  1916, 
By  the  MACMILLAN  COMPANY. 


Set  up  and  electrotyped.     Published  September,  1916.     Reprinted 
March,  1917. 


3 


NoriDDDli  iPrfgg 

J.  S.  dishing  Co.  —  Berwick  &  Smith  Co, 

Norwood,  Mass.,  U.S.A. 


w^    ^A 


PREFACE 

This  book  presents  a  first  course  in  the  calculus  sub- 
stantially as  the  author  has  taught  it  at  the  University  of 
Michigan  for  a  number  of  years.  The  following  points 
may  be  mentioned  as  more  or  less  prominent  features  of 
the  book.  * 

In  the  treatment  of  each  topic,  the  text  is  intended  to 
contain  a  precise  statement  of  the  fundamental  principle 
involved,  and  to  insure  the  student's  clear  understanding 
of  this  principle,  without  distracting  his  attention  by  the 
discussion  of  a  multitude  of  details.  The  accompanying 
exercises  are  intended  to  present  the  problem  in  hand  in 
a  great  variety  of  forms  and  guises,  and  to  train  the  stu- 
dent in  adapting  the  general  methods  of  the  text  to  fit 
these  various  forms.  The  constant  aim  is  to  prevent  the 
work  from  degenerating  into  mere  mechanical  routine,  as 
it  so  often  tends  to  do.  Wherever  possible,  except  in  the 
purely  formal  parts  of  the  course,  the  summarizing  of  the 
theory  into  rules  or  formulas  which  can  be  applied  blindly 
has  been  avoided.  For  instance,  in  the  chapter  on  geo- 
metric applications  of  the  definite  integral,  stress  is  laid 
on  the  fact  that  the  basic  formulas  are  those  of  elemen- 
tary geometry,  and  special  formulas  involving  a  coordinate 
system  are  omitted. 

Where  the  passage  from  theory  to  practice  would  be 
too  difficult  for  the  average  student,  worked  examples  are 
inserted. 

It  seems  clear  that  so-called  applications  in  which  the 
student  is  made  to  use  a  formula  without  explanation  of 


•laa^Q'T 


vi  PREFACE 

its  meaning  and  derivation,  are  of  little  value.  In  the 
present  text  the  non-geometric  applications  are  taken  sys- 
tematically from  one  subject,  mechanics,  and  the  theory 
is  developed  as  fully  as  in  the  calculus  proper. 

A  feature  of  the  book  is  its  insistence  on  the  importance 
of  checking  the  results  of  exercises,  either  directly  or  by 
solving  in  more  than  one  way.  The  latter  method  is 
largely  used  in  the  integral  calculus,  on  account  of  the 
variety  of  elementary  transformations  possible  with  defi- 
nite integrals. 

The  answers  to  many  of  the  exercises  are  given,  but 
seldom  where  a  knowledge  of  the  answer  would  help  in 
the  solution,  or  where  a  simple  means  of  checking  the 
answer  exists. 

Topics  of  minor  importance  are  presented  in  such  a  way 
that  they  may  be  omitted  if  it  is  desired  to  give  a  short 
course. 

The  chapter  on  curve  tracing  is  introduced  as  early  as 
possible,  so  that  the  results  are  available  for  use  through- 
out the  course. 

Some  instructors  will  wish  to  begin  the  use  of  integral 
tables  immediately  after  the  chapters  on  formal  integra- 
tion.    This  of  course  can  easily  be  done. 

In  spite  of  obvious  difficulties,  a  chapter  embodying  a 
first  treatment  of  centroids  and  moments  of  inertia  is 
introduced  before  multiple  integrals  have  been  defined. 
By  this  arrangement  the  student  is  brought  to  realize 
the  fact  that  in  most  cases  of  practical  importance  mass- 
moments  of  the  first  and  second  orders  can  be  found  by 
simple  integration,  whereas  from  the  usual  treatment  he 
gets  exactly  the  opposite  idea. 

In  the  chapters  on  differential  equations,  emphasis  is 
laid  on  those  types  most  likely  to  be  met  by  the  student 
of  engineering  or  the  mathematical  sciences.  In  the  last 
chapter  the  average  student  will  doubtless  require  con- 


PREFACE  vii 

siderable  help  from  the  instructor,  but  it  is  hoped  that, 
if  properly  presented,  the  chapter  may  give  the  student 
some  facility  in  writing  and  solving  the  simpler  differen- 
tial equations  of  mechanics  and  in  interpreting  the  results. 
To  Professor  Alexander  Ziwet,  who  has  read  the  entire 
manuscript,  the  author  makes  grateful  acknowledgment, 
not  only  for  valuable  advice  and  criticism,  but  for  his 
unfailing  encouragement  and  support.  Thanks  are  also 
due  to  Professor  T.  H.  Hildebrandt,  who  has  kindly 
assisted  in  reading  the  proofs,  and  has  made  a  number  of 
useful  suggestions. 

CLYDE   E.  LOVE. 

Ann  Arbor, 

August,  1916. 


CONTENTS 

CHAPTER   I 
FUNCTIONS.     LIMITS.     CONTINUITY 

ART,  PAGK 

1.  Functions 1 

2.  Geometric  representation        .......  2 

3.  Independent  variable      ........  3 

4.  Kinds  of  functions 3 

5.  One-valued  and  many-valued  functions           ....  3 

6.  Rate  of  change  ;  slope 5 

7.  Limits 6 

8.  Theorems  on  limits 7 

9.  Limit  of  a  function          ........  8 

10.  Infinitesimals ..........  8 

11.  Limit  of  the  ratio  of  two  infinitesimals          ....  8 

12.  Continuity 10 

13.  Infinity 11 

14.  Function  with  infinite  argument 12 

CHAPTER   II 

THE   DERIVATIVE 

15.  The  derivative 14 

16.  Higher  derivatives .  18 

CHAPTER   III 

DIFFERENTIATION   OF   ALGEBRAIC   FUNCTIONS 

17.  Introduction 19 

18.  Derivative  of  a  constant          .......  19 

19.  Derivative  of  a  sum  ;  a  product ;  a  quotient           ...  19 

20.  Derivative  of  a  function  of  a  function 21 

ix 


CONTENTS 


21.  Derivative  of  x^,  n  a  positive  integer 

22.  Derivative  of  x",  n  fractional 

23.  The  general  power  formula    . 

24.  Implicit  functions  .... 

25.  Differentiation  of  implicit  functions 

26.  Inverse  functions     .         ... 


PAGE 

22 
24 
24 
26 
26 
27 


CHAPTER   IV 

GEOMETRIC   APPLICATIONS 

27.  Tangents  and  normals  to  curves -.29 

28.  Length  of  tangent,  subtangent,  normal,  and  subnormal        .  30 

29.  Increasing  and  decreasing  functions       .....  32 

30.  Maxima  and  minima 33 

31.  Concavity 33 

32.  Points  of  inflection 35 

'33.    Summary  of  tests  for  maxima  and  minima,  etc.    ...  35 

34.  Applications  of  maxima  and  minima 37 

35.  Derived  curves 42 


CHAPTER   V 
DIFFERENTIATION  OF  TRANSCENDENTAL  FUNCTIONS 


I.     Trigonometric  and  Inverse  Trigonometric  Functions 


36.  Trigonometric  functions 

37.  Differentiation  of  sin  x   , 

38.  Limit  of  sin  a/ a  as  ct  approaches  0 

39.  Differentiation  of  cos  x,  tan  x,  etc. 

40.  Inverse  trigonometric  functions 

41.  Restriction  to  a  single  branch 

42.  Differentiation  of  the  inverse  trigonometric  functions 


II.     Exponential  and  Logarithmic  Functions 

43.  Exponentials  and  logarithms  .    -     . 

44.  Properties  of  logarithms 

45.  The  derivative  of  the  logarithm 

46.  The  limit  e 

47.  Differentiation  of  the  exponential  function 


48.    Hyperbolic  functions 


45 
46 
47 
48 
51 
51 
53 


55 
57 
58 
60 
62 
64 


CONTENTS  xi 

CHAPTER   VI 
THE  DIFFERENTIAL 

ART.  PAGE 

r  49.   Order  of  infinitesimals 68 

\    50.    The  differential 69 

\51.    Parametric  equations ;  implicit  functions       ....  72 

CHAPTER   VII 

CURVATURE 

52.  Differential  of  arc 75 

53.  Curvature 76 

54.  Radius  of  curvature 78 


CHAPTER   VIIT 

APPLICATIONS   OF   THE   DERIVATIVE  IN 

MECHANICS 

//^  .  .... 

jf55.   Velocity  and  acceleration  in  rectilinear  motion     ...       80 

56.  Vectors     ...  82 

57.  Velocity  in  curvilinear  motion 82 

58.  Rotation ...........  83 

59.  Acceleration  in  curvilinear  motion  .....  85 

60.  Time-rates 88 


■/S- 


*v 


\ 


CHAPTER   IX 


t         CURVE   TRACING  IN   CARTESIAN   COORDINATES 

I.     Algebraic  Curves 

61.  Introduction 92 

62.  Singular  points        .........  92 

63.  Determination  of  tangents  by  inspection        .         .         .         .  93 

64.  Kinds  of  singular  points          .......  95 

65.  Asymptotes 97 

66.  Exceptional  cases    .........  101 

67.  General  directions  for  tracing  algebraic  curves      .         .         .  101 


xu 


CONTENTS 


II.    Transcendental  Curves 


ART 

68. 
69. 
70. 
71. 


Tracing  of  transcendental  curves  . 
Curve  tracing  by  composition  of  ordinates 
Graphic  solution  of  equations 
The  cycloid 

72.  The  epicycloid 

73.  The  hypocycloid 


PAGE 

105 
106 
107 
108 
109 
110 


CHAPTER   X 

CURVE   TRACING  IN   POLAR,  COORDINATES 

74.  Slope  of  a  curve  in  polar  coordinates     .... 

75.  Maxima  and  minima 

76.  Curve  tracing 


112 
114 
114 


CHAPTER  XI 


THE   INDEFINITE  INTEGRAL 


77.  Integration 

78.  Integration  an  indirect  process 

79.  Constant  of  integration  .         .         .         . 

80.  Functions  having  the  same  derivative   . 

81.  Geometric  interpretation  of  an  integral 

82.  Variable  of  integration  .         .         .         , 

83.  Change  of  the  variable  of  integration    . 

84.  Integration  by  substitution     . 


116 
118 
118 
119 
120 
122 
122 
123 


CHAPTER   XII 

STANDARD   FORMULAS  OF  INTEGRATION 

85.  Standard  formulas 

86.  Formulas  (l)-(3) 

87.  Formula  (4) :  Powers . 

88.  Formulas  (5) -(6)  :  Logarithms  and  exponentials 

89.  Formulas  (7)-(9)  :  Trigonometric  functions 

90.  Formulas  (lO)-(ll)  :  Inverse  trigonometric  functions 

91.  Formula  (12)  :  Integration  by  parts      .... 

92.  Integration  by  substitution     ...... 


126 
127 
127 
129 
131 
132 
132 
134 


CONTENTS  xiil 

CHAPTER   XIII 
INTEGRATION   OF   RATIONAL   FRACTIONS 

ART.  PAGE 

93.  Preliminary  step 137 

94.  Partial  fractions 137 

95.  Distinct  linear  factors 137 

96.  Repeated  linear  factors 139 

97.  Quadratic  factors 140 


CHAPTER   XIV 

THE     DEFINITE   INTEGRAL 

98.  The  definite  integral 143 

99.  Geometric  interpretation  of  a  definite  integral    .         .         .  144 

100.  Interchanging  limits 145 

101.  Change  of  limits  corresponding  to  a  change  of  variable      .  145 


CHAPTER   XV 

THE   DEFINITE   INTEGRAL   AS   THE   LIMIT   OF 

A  SUM 

102.    Area  under  a  curve 148  * 

108.   Evaluation  of  the  limit 150  • 

104.  The  fundamental  theorem 150  •• 

105.  Plane  areas  in  cartesian  coordinates     .....  151  • 

106.  Plane  areas  in  polar  coordinates  .         .         .         .         .         .154 

107.  Volumes  of  revolution 156 

108.  Volumes  of  revolution  :  second  method        ....  157 

109.  A  theorem  on  infinitesimals 158 

110.  Other  volumes 161 

111.  Line  integrals 163 

112.  Geometric  interpretation  of  the  line  integral       .         .         .  164 

113.  Fundamental  theorem  for  line  integrals       ....  165 

114.  Evaluation  of  line  integrals  .         ......  165 

115.  Length  of  a  curvilinear  arc  .......  167 

116.  Surfaces  of  revolution   ........  168 

117.  Cylindrical  surfaces 170 


xiv  '  CONTENTS 


CHAPTER   XVI 
INTEGRAL   TABLES 

ART.  PAGE 

118.   Use  of  tables ,        .        .    172 


CHAPTER   XVII 

IMPROPER  INTEGRALS 

119.  Definitions 175 

120.  Geometric  interpretation 177 

CHAPTER   XVIII 

CENTROIDS.     MOMENTS   OF   INERTIA 

I.     Centroids 

121.  Mass;  density        .         .         .         ...         .        .         .         .179 

122.  Moment  of  mass 180 

123.  Centroid •     .  180 

124.  Centroids  of  geometrical  figures  .         .         .         .         .         .  182 

125.  Determination  of  centroids  by  integration  ....  183 

126.  Centroids  of  plane  areas 185 

127.  Centroids  of  volumes 187 

128.  Centroids  of  lines 188 

129.  Centroids  of  curved  surfaces 189 

II.     Moments  of  Inertia 

130.  Moment  of  inertia 190 

131.  Radius  of  gyration 190 

132.  Determination  of  moment  of  inertia  by  integration    .         .  190 

133.  Moment  of  inertia  with  respect  to  a  plane  ....  194 

134.  General  theorems  on  moments  of  inertia     .         .         .         .,  195 

135.  Kinetic  energy  of  a  rotating  body         .         .        .        »        .  198 

CHAPTER   XIX 

LAW   OF   THE  MEAN.     EVALUATION   OF   LIMITS 

136.  Rolle's  theorem 200 

137.  The  law  of  the  mean     .         .         .         .       '.         .         .         .200 

138.  Other  forms  of  the  law  of  the  mean 201 


CONTENTS 


XV 


ART. 

139.  The  indeterminate  forms  -,  —     . 

0    00 

140.  The  indeterminate  forms  0  •  go,  co  —  oo 

141.  General  remarks  on  evaluation  of  limits 


PAGE 

202 

204 
206 


CHAPTER   XX 

INFINITE   SERIES.     TAYLOR'S   THEOREM 
I.     Series  of  Constant  Terms 

142.  Series  of  n  terms 209 

143.  Infinite  series         . 209 

144.  Sum  of  an  infinite  series 210 

145.  Convergence  and  divergence          .       ' 211 

146.  Tests  for  convergence    .         .         .        .         .        .         .         .211 

147.  Cauchy's  integral  test 212 

148.  Comparison  test 214 

149.  Ratio  test 216 

150.  Alternating  series 218 

151.  Absolute  convergence 219 


II.     Power  Series 

152.  Power  series  . 

153.  Maclanrin's  series . 

154.  Taylor's  series 
X|(^55.  Taylor's  theorem  . 

"^^56.  Approximate  computation  by  seiies 

157.  Operations  with  power  series 

158.  Computation  of  logarithms  . 


220 
222 
223 
226 
228 
230 
234 


CHAPTER   XXI 

FUNCTIONS   OF   SEVERAL   VARIABLES 
I.     Partial  Differentiation 

159.  Functions  of  several  variables 

160.  Limits;  continuity        ....... 

161.  Partial  derivatives 


236 
236 
237 


XVI 


CONTENTS 


162.  Geometric  interpretation  of  partial  derivatives 

163.  Higher  derivatives         ..... 

164.  Total  differentials 

165.  Differentiation  of  implicit  functions    . 


PAGE 

238 
238 
240 
241 


II.     Applications  to  Solid  Analytic  Geometry 

166.  Tangent  plane  to  a  surface 244 

167.  Normal  line  to  a  surface 245 

168.  Angle  between  two  surfaces ;  between  a  line  and  a  surface  246 

169.  Space  curves 248 

170.  Tangent  line  and  normal  plane  to  a  space  curve  .         .  248 

171.  Direction  cosines  of  the  tangent 250 

172.  Length  of  a  space  curve 250 


CHAPTER   XXII 


ENVELOPES.     EVOLUTES 

173.  Envelope  of  a  family  of  plane  curves  . 

174.  Determination  of  the  envelope     . 

175.  Envelope  of  tangents     ..... 

176.  The  evolute 


252 
252 
254 

256 


CHAPTER   XXIII 


MULTIPLE   INTEGRALS 

177.  Volume  under  a  surface 

178.  Volume  under  a  surface  :  second  method 

179.  Interpretation  of  the  given  function    . 

180.  The  double  integral       .... 

181.  The  double  integral  in  polar  coordinates 

182.  Transformation  of  double  integrals 

183.  Area  of  a  surface 

184.  Triple  integrals 

185.  Heterogeneous  masses  .... 

186.  Centroids  and  moments  of  inertia  :  the  general  case 


258 
262 
263 
264 

265 
266 
268 
270 
273 
276 


CONTENTS 


xvii 


ART. 

187. 
188. 
189. 
190. 


CHAPTER  XXIV 
FLUID  PRESSURE 


Force 

Force  distributed  over  an  area 

Fluid  pressure 

Resultant  of  parallel  forces  . 


191.   Center  of  pressure 


PAGE 

279 

280 
281 
283 

284 


CHAPTER  XXY 
DIFFERENTIAL  EQUATIONS  OF  THE  FIRST  ORDER 


I.     General  Introduction 


192.  Differential  equations    . 

193.  Order  of  a  differential  equation    . 

194.  Solutions  of  a  differential  equation 


286 
287 
287 


II.     Equations  of  the  First  Order 

195.  The  general  solution 287 

196.  Particular  solutions       ........  289 

197.  Geometrical  interpretation 290 

198.  Separation  of  variables 292 

199.  Coefficients  homogeneous  of  the  same  degree       .         .         .  294 

200.  Exact  differentials .295 

201.  Exact  differential  equations 296 

202.  Integrating  factors 297 

203.  The  linear  equation 298 

204.  Equations  linear  in/ (j/) 300 

205.  Geometric  applications ,  301 


CHAPTER  XXVI 

DIFFERENTIAL   EQUATIONS   OF   HIGHER   ORDER 
I.     Introduction 

206.  General  and  particular  solutions  .         .         .         .         .         .     304 

207.  Geometric  interpretation 305 


XVlll 


CONTENTS 


11.     The  Linear  Equation  with  Constant  Coefficients 


208.  The  linear  equation       .... 

209.  The  homogeneous  linear  equation 

210.  The  characteristic  equation  .         .       '  . 

211.  Distinct  roots 

212.  Repeated  roots       .....' 

213.  Complex  roots        ..... 

214.  Extension  to  equations  of  higher  order 

215.  The  non -homogeneous  linear  equation 


307 
307 
308 
309 
309 
310 
312 
313 


III.     Miscellaneous  Equations  of  the  Second  Order 

216.  The  equation  ?/"= /(^) 316 

217.  The  equation  y"  =  /{y) 317 

218.  Dependent  variable  absent 318 

219.  Independent  variable  absent 319 


CHAPTER   XXVII 

APPLICATIONS   OF    DIFFERENTIAL   EQUATIONS 
IN   MECHANICS 

I.     Rectilinear  Motion 


220.  Rectilinear  motion         .... 

221.  Motion  of  a  particle  under  given  forces 

222.  The  equation  of  motion 

223.  Uniformly  accelerated  motion 

224.  Momentum  ;  impulse    .... 

225.  The  principle  of  impulse  and  momentum 

226.  Work 

227.  The  principle  of  kinetic  energy  and  work 

228.  Constrained  motion       .... 

229.  Simple  harmonic  motion 

230.  Attraction  proportional  to  the  distance 

231.  Hooke's  law 


321 
322 
323 
324 
326 
326 
327 
328 
329 
330 
332 
333 


11.     Plane  Curvilinear  Motion 

232.  Rotation 335 

233.  The  simple  pendulum    ........  336 

234.  The  equations  of  motion        .         .         .         .         .         .         .  337 

235.  Projectiles      .         . 338 


DIFFERENTIAL   AND    INTEGRAL    CALCULUS 


CALCULUS 

CHAPTER  I 
FUNCTIONS.     LIMITS.     CONTINUITY 

1.  Functions.  If  a  variable  y  depends  upon  a  variable 
X  so  that  to  every  value  of  x  there  corresponds  a  value  of 
y,  then  y  is  said  to  be  ?,  function  of  x. 

For  example,  (a)  the  area  of  a  circle  is  determined  by 
the  radius  and  is  therefore  a  function  of  the  radius ; 
(S)  the  attraction  (or  repulsion)  between  two  magnetic 
poles  is  a  function  of  the  distance  between  them ;  (c)  the 
volume  of  a  given  mass  of  gas  at  a  constant  temperature  is 
a  function  of  the  pressure  upon  the  gas. 

A  complete  study  of  the  properties  of  a  function  is 
possible  in  general  only  when  the  function  is  given  by  a 
definite  mathematical  expression.  For  this  reason  we 
shall  be  concerned  almost  entirely  with  functions  defined 
in  this  way.  Thus,  in  the  examples  above,  we  have 
(a)  A  =  7rr2, 

(c)  for  a  "perfect  gas," 

h 
v  =  -. 

V 

But  the  existence  of  a  functional  relation  between  two 

quantities  does  not  imply  the  possibility  of  giving  this 
relation  a  mathematical  formulation.  If  by  any  means 
whatever   a   value  of  y  is  determined   corresponding  to 

B  1 


2  CALCULUS 

everj  ^^a«.ue  of  x  und'3^  consideration,  then  y  is  a  function 
of  X.  For  example,  the  temperature  of  the  air  at  any 
point  of  the  earth's  surface  is  a  function  of  the  time  at 
which  the  thermometer  is  read,  although  no  mathematical 
law  connecting  the  two  variables  is  known. 

We  often  wish  to  express  merely  the  fact  that  «/  is  a 
function  of  x^  without  assigning  the  particular  form  of  the 
function.     This  is  done  by  writing 

y  =/(^) 
(read  y  equals  /  of  a;).     Other  letters  may  of  course  be 
used  in  the  functional  symbol,  as  F{x)^  </>(^)i  '^C^)?  ^tc. 

The  value  of  f(x)  when  x  =  ai^  denoted  by  the  symbol 
/(a).     Thus,  if 

f(x)=x^-Zx-l, 
then 

/(a)  =  a2  -  3  a  -  1, 
/(2)  =  -  3, 

f(^x  +  ii)  =  {x  +  hy  -  s(x  +  A)- 1. 

Except  where  the  contrary  is  explicitly  stated,  the  vari- 
ables and  functions  with  which  we  shall  have  to  deal  are 
restricted  to  real  values.  This  restriction  is  introduced 
for  the  sake  of  simplicity,  and  also  because  in  the  elemen- 
tary applications  only  real  quantities  are  of  importance. 

2.  Geometric  representation.  The  student  is  already 
familiar  with  the  geometric  representation  of  a  function 
as  the  ordinate  of  a  plane  curve.  Thus  in  (a)  of  §  1  the 
graph  is  a  parabola  ;  in  (5)  it  is  a  certain  cubic  curve;  in 
(c)  it  is  an  equilateral  hyperbola. 

Even  though  no  mathematical  expression  for  the  func- 
tion is  known,  it  may  still  be  represented  graphically.  For 
instance,  to  represent  the  temperature  at  a  point  of  the 
earth's  surface  as  a  function  of  the  time,  let  a  large  number 
of  readings  be  taken,  the  corresponding  points  be  plotted  on 
coordinate  paper  with  time  as  abscissa  and  temperature  as 


FUNCTIONS.     LIMITS.     CONTINUITY  3 

ordinate,  and  a  smooth  curve  be  drawn  through  the  points. 
This  curve  will  represent  approximately  the  variation  of 
temperature  throughout  the  time-interval  in  question. 

3.  Independent  variable.  We  usually  think  of  x  as 
varying  arbitrarily  —  i.e.  we  assign  values  to  x  at  pleasure, 
and  compute  the  corresponding  values  of  y.  The  variable 
X  is  then  called  the  independent  variable^  or  argument.  But 
it  is  clear  that  if  y  is  a  function  of  x^  x  is  likewise  a  func- 
tion of  y,  and  in  general  either  one  may  be  chosen  as  the 
independent  variable. 

The  values  assigned  to  x  must  of  course  be  compatible 
with  the  conditions  of  the  problem  in  hand.  In  most  cases 
X  is  restricted  to  a  definite  range  or  interval ;  for  instance, 
if  the  function  we  are  dealing  with  is  «/  =  V2;,  we  restrict 
X  to  positive  values. 

4.  Kinds  of  functions.  We  shall  have  to  deal  with  both 
algebraic  and  transcendental  functions.  The  algebraic 
functions  are  rational  integral  functions^  or  polynomials ; 
rational  fractions^  or  quotients  of  polynomials ;  and  irra- 
tional functions,  of  which  the  simplest  are  those  formed 
from  rational  functions  by  the  extraction  of  roots.  The 
elementary  transcendental  functions  are  trigonometric  and 
inverse  trigonometric  functions  ;  exponential  functions.,  in 
which  the  variable  occurs  as  an  exponent;   and  logarithms. 

Function 


algebraic  transcendental 


rational  irrational  elementary  higher 


integral  fractional  trigonometric  exponential 

inverse  trig'c  logarithmic 

5.  One-valued  and  many-valued  functions.  A  function 
y  =f(x')  is  said  to  be  one-valued,  if  to  every  value  of  x 
corresponds  a  single  value  of  y ;  two-valued,  if  to  every 
value  of  X  correspond  two  values  of  y,  etc. 

In  the  case  of    a  many-valued  function  it  is  usual  to 


4  CALCULUS 

group  the  values  in  such  a  way  as  to  form  a  number  of  one- 
valued  functions,  called  the  branches  of  the  original  func- 
tion.    Thus  the  equation 

defines  a  two-valued  function  whose  branches  are 

y  =  —  '^x. 
In  dealing  with  many- valued  functions,  we  shall  in  gen- 
eral confine  our  attention  to  a  particular  branch. 

/ 

EXERCISES  , 

1.  Express  the  surface  and  volume  of  a  sphere  as  functions  of  the 
radius ;  the  radius  as  a  function  of  the  surface  and  of  the  volume. 

2.  Express  the  surface  and  volume  of  a  cube  as  functions  of  the 
length  of  its  edge. 

3.  Represent  geometrically  each  of  the  functions  of  Ex.  2. 

4.  rind/(0,/(3),/(-l),/(0),/(:r  +  A),if 

(a)  f{x)  =  2  a:  +  5 ;  {h)  f{x)  =  x^  -  3  x  +  3  ; 

(c)/(x)  =  sin7rx;  {d)  f{x)  =  2\ 

5.  Exhibit  graphically  each  of  the  functions  of  Ex.  4. 

6.  Plot  the  graph  of  each  of  the  functions  (a),  (6),  (c)  of  §  1. 

7.  Restate  the  examples  (a),  (h),  (^)  of  §  1  both  in  words  and  by 
an  equation,  with  the  independent  and  the  dependent  variable  inter- 
changed. 

8.  Plot  the  graph  of  each  of  the  following  functions: 

l-\-  x^ 

(0  y  =  7-^2'  ("^^  y  =  r-^' 

1  +   X^  1   —  X 

f 

9.  Show  that  (a)  the  graph  of  a  one-valued  function  is  met  by 
any  parallel  to  the  y-axis  in  not  more  than  one  point ;  (b)  the  graph 
of  a  many-valued  function  consists  of  a  number  of  branches  (not 
necessarily  disconnected),  each  of  which  has  this  same  property. 
Give  examples. 

10.  Show  that  the  equation  i/^  =  x'^  —  a^  defines  y  as  a  two-valued 
function  of  x,  and  draw  the  graph. 


FUNCTIONS.      LIMITS.      CONTINUITY 


11.  The  freezing  point  of  water  is  32°  Fahrenheit,  0°  Centigrade ; 
the  boiling  point,  212^  F.,  100°  C.  Express  temperature  in  degrees 
F.  as  a  function  of  temperature  in  degrees  C,  both  analytically  and 
graphically. 

12.  A  sum  of  money  is  placed  at  simple  interest.  Express 
the  amount  at  any  time  as  a  function  of  the  time,  and  draw  the 
graph.  , 

6.  Rate  of  change ;  slope.  A  fundamental  problem  in 
studying  the  nature  of  a  function  is  the  determination  of 
its  rate  of  change. 

Let  P  :  (x^  y)  be  a  point  on  the  graph  of  the  function 

Assign  to  X  an  arbitrary  change,  or  increment^  Ax  (read 
delta  ic,  not  delta  times  a:),  usually  taken  positive,  and 
denote  by  Ay  the  corresponding 
change  in  ?/,  so  that  the  point 
P'  :  (x  -{-  Aa:,  y  +  Ay)  is  a  second 
point  on  the  curve.     The  ratio 


Fig.  1 


— ^  is  the  average  rate  of  change 

Ax 

of  g  with  respect  to  x  in  the  in- 
terval Ax;  geometrically  this 
ratio  is  the  slope  of  the  chord 

PP',     If  now  we  let  Ax  approach  0,  the  ratio  — ^  in  gen- 

Ax 

eral  approaches  a  definite  limiting  value,  which  is  de- 
fined as  the  rate  of  change  of  g  with  respect  to  x  at  the 
point  P. 

The  geometric  interpretation  is  obvious  :  when  Ax  is 
taken  smaller  and  smaller,  P'  approaches  P  along  the 
curve,  the  chord  PP'  approaches  the  tangent  at  P  as  its 

limiting  position,  and  —^  approaches  as  its  limit  the  slope 

of  the  tangent.  Hence  the  rate  of  change  of  a  function  is 
the  slope  of  its  graph. 


6  CALCULUS 

7.  Limits.  From  what  has  just  been  said,  it  appears 
that  the  determination  of  the  rate  of  change  of  a  function, 
or  the  slope  of  a  curve,  requires  the  evaluation  of  a  certain 
limit.  It  will  therefore  be  well  to  introduce  at  this  point 
a  brief  discussion  of  the  subject  of  limits. 

When  the  successive  values  of  a  variable  x  approach 
nearer  and  nearer  a  fixed  number  a,  in  such  a  way  that  the 
difference  a  —  x  becomes  and  remains  numerically  less  than 
any  preassigned  positive  number  however  small,  the  con- 
stant a  is  called  the  limit  of  a:,  and  x  is  said  to  approach  the 
limit  a  —  in  symbols, 

lim  X  =  a. 

Examples  are  easily  found  in  elementary  work : 

(a)  If  a  regular  polygon  be  inscribed  in  a  circle,  the 
difference  between  the  area  Ap  of  the  polygon  and  the  area 
Ac  oi  the  circle  becomes  arbitrarily  small  (less  than  any 
preassigned  number)  as  the  number  of  sides  increases  in- 
definitely.    Hence 

Irm  Ap  =  Ac 

(6)  We  know  from  elementary  algebra  that  the  sum 
Sn  of  the  geometric  series 


IS 


+ 

1 

2n-l 

2- 

1 
2n-l 

is 

1 

1-i 

The  difference  between  2  and  S^  is 

2n-l 

This  difference  becomes  arbitrarily  small  as  the  number  of 
terms  increases  indefinitely  ;   hence 

lim  S,,  =  2. 
((?)  If  a  steel  spring  of  length  I  suspended  vertically  be 
stretched  to  a  length  I  -{-  a  and  then  released,  tlie  end  of 


FUNCTIONS.     LIMITS.     CONTINUITY  7 

the  spring  will  oscillate  about  its  original  position. 
The  length  x  of  the  spring  will  be  alternately  greater  and 
less  than  the  original  length  Z,  but  as  the  oscillations  be- 
come smaller  the  difference  between  x  and  I  will  become 
and  remain  arbitrarily  small.     Thus 

lim  x  —  l. 
In  this  example,  the  variable  actually  reaches  its  limit, 
since  the  spring  soon  ceases  to  oscillate  at  all.  In  many 
cases,  however,  the  variable  never  reaches  its  limit. 
This  is  true  in  (a)  above,  since  no  matter  how  many 
sides  the  polygon  may  have,  its  area  is  always  less  than 
that  of  the  circle. 

8.  Theorems  on  limits.  We  shall  have  occasion  to  use 
the  following  theorems  on  limits,  which  we  assunae  without 
formal  proof. 

Theorem  I* :  The  limit  of  the  sum  of  two  variables  is  equal 
to  the  sum  of  their  limits. 

Theorem  II :  The  limit  of  the  product  of  two  variables 
is  equal  to  the  product  of  their  limits. 

Theorem  III:  The  limit  of  the  quotient  of  two  variables 
is  equal  to  the  quotient  of  their  limits^  provided  the  limit  of 

the  denominator  is  not  0. 

f  increases  i 
Theorem  IV :  If  a  variable  steadily    \  ,  \    but 

I  decreases  j 

never  becomes     .  \  than  some  fixed  number  A.,  the  vari- 

able approaches  a  limit  which  is  not  \  \  than  A. 

Theorems  I  and  II  may  evidently  be  extended  to  the 
case  of  any  number  of  variables. 

*  In  theorems  I,  II,  III  it  is  of  course  implied  that  the  limits  of  the  two 
variables  exist.  We  shall  see  later  (§§  139,  140)  that  the  sum  of  two  vari- 
ables, for  instance,  may  approach  a  limit  when  neither  of  the  two  variables 
taken  by  itself  approaches  a  limit. 


8  CALCULUS 

9.  Limit  of  a  function.  We  have  frequently  to  observe 
the  behayior  of  a  function  f(x)  as  the  argument  x  ap- 
proaches a  limit.  If,  as  x  approaches  a,  the  difference  be- 
tween f(x)  and  some  fixed  number  I  ultimately  becomes 
and  remains  numerically  less  than  any  preassigned  constant 
however  small,  the  function  f(x)  is  said  to  approach  the 
limit  Z,  and  we  write 

Unless  otherwise  specified  it  is  supposed  that  the  same 
limit  is  approached  whether  x  comes  up  to  a  from  the 
positive  or  the  negative  direction.  If  we  wish  to  consider 
what  happens  when  x  approaches  a  from  the  positive  side 

only,  we  write     ^^\  /(^) ;    from  the  negative  side  only, 
'""  fix). 

10.  Infinitesimals.  An  infinitesimal  is  a  variable  whose 
limit  is  0.  Thus  a  constant,  however  small,  is  not  an  in- 
finitesimal. An  infinitesimal  is  not  necessarily  small  at 
all  stages  of  its  variation ;  the  only  thing  necessary  is  that 
ultimately  it  must  become  and  remain  numerically  less  than 
any  assignable  constant  however  small. 

If  one  infinitesimal  is  a  function  of  another,  the  inde- 
pendent variable  is  called  the  principal  infinitesimal. 

In  the  problem  of  §  6,  both  Aa;  and  A^  are  infinitesimals, 
with  ^x  as  the  principal  infinitesimal. 

11.  Limit  of  the  ratio  of  two  infinitesimals.  We  return 
to  the  exceptional  case  of  theorem  III,  §  8,  in  which  the 

denominator  is  infinitesimal.       Given  any  fraction   -   in 

which  u  approaches  0,  two  cases  are  to  be  distinguished : 
(a)  V  also  approaches  0  ; 
(J)  V  does  not  approach  0. 

It  is  clear  that  in  case  (6)  the  fraction  —  may  be  made 

u 

to  assume  values  greater  than  any  assignable  constant  by 


FUNCTIONS.     LIMITS.     CONTINUITY  9 

taking  u  sufficiently  small ;  hence  the  fraction  can  ap- 
proach no  limit.  But  consider  case  (a),  in  which  both  u 
and  V  are  infinitesimal.     Theorem  III  does  not  apply;   the 

ratio  of  the  limits  is  -,  which  is  quite  meaningless  ;  never- 
theless the  limit  of  the  ratio  may  exist,  as  we  shall  find  in 
many  cases  in  the  next  few  chaj)ters. 

The  determination  of  the  limit  of  the  ratio  of  two 
infinitesimals  is  a  problem  of  the  greatest  importance ; 
in  fact,  it  is  clear  from  the  discussion  of  §  6  that  this 
problem  always  arises  in  finding  the  rate  of  change  of  a 
function,  or  the  slope  of  a  curve. 

EXERCISES 

1.  Determine  (a)     1"^  (x^  -  3  x^  -  5  a:  -  5)  ; 

"-  ^    x>— 1 

/j\    lim  ^   —  X  —  \ 

Which  of  the  theorems  of  §  8  are  needed? 

2.  Determine  (a)   ^i^]  x-^  -  3  a;  +  2 

(b)  l^^l  (sin  X  +  cos  x)  ; 

(c)  lira ^.  Ans.  (c)    -2 


2x 
Which  of  the  theorems  of  §  8  are  needed  ? 

3.  Determine    lim  x'^  -  3  .r  +  2      ^y^ich  of    the   theorems   of   §  8 

•^■>l        x  -1 
are  used?  Ans.  —  1. 

4.  Evaluate  ^i"^     ^  ~  ^'  • 


5.   Evaluate 


lim  Vl  -  x\  ^^^^  1  Vg 


6.  Evaluate  J^^^^^-  ^ns.  1. 

■^^■^tan  X 

7.  Evaluate  ^i^  sin_2£.  Ans.  2. 

•^■^0  tan  X 


10  CALCULUS 

8.  Evaluate  ^"^  ^^^Ix 

•^■^0  sin  X 

9.  Show  that,  if  n  is  a  positive  integer, 

lim  x""  =  (lini  xy. 

10.  Show  that,  if  P{x)  is  a  polynomial  in  x, 

11.  Show  that,  if  P^ix)  and  P2(x)  are  polynomials, 

lim^iM  =  ^PiOO 
^>«  PaCa:)       P2(a)' 
provided  PgC*^)  =^  ^^ 

12.  Under  what  circumstances  may  the  limit  in  Ex.  11  exist  when 
P2(a)  =  0  ?     Give  an  example. 

13.  Does  the  limit  in  Ex.  11  always  exist  when  Pi(a)  =  P2(a)  =  0? 
Give  examples. 

12.  Continuity.  An  important  idea  in  the  study  of 
functions  is  that  of  continuity. 

A  function  /(re)  is  said  to  be  continuous  at  the  point 
a;  =  a  if 

This  means,  first,  that  the  function  is  jigfined  when  x  =  a^ 
and  second,  that  the  difference  between  /(a:)  and  /(a)  be- 
comes and  remains  arbitrarily  small  (numerically  less  than 
any  assignable  constant)  as  x  approaches  a.  The  curve 
y  =f(x')  passes  through  the  point  x  =  a  without  a  gap  or 
break. 

A  function  is  said  to  be  continuous  in  an  interval  of 
values  of  the  argument  if  it  is  continuous  at  all  points  of 
the  interval.   ^ 

In  the  discussion  of  §  6,  it  is  ta_citly  assumed  that  the 
function  is  continuous  in  an  interval  including  the  point 
P\  this  assumption  is  an  essential  part  of  the  argument. 

All  the  functions  treated  in  this  hook  are  continuous.,  ex- 
cept perhaps  for  certain  particular  values  of  the  variable, 
and  such  values  are  either  excluded  or  subjected  to  special 
investigation. 


FUNCTIONS.     LIMITS.     CONTINUITY 


11 


13.  Infinity.  The  most  important  type  of  discontinuity 
is  that  in  which  the  function  increases  numerically  without 
limit,  or,  as  we  say,  becomes  infinite^  as  x  approaches  a.  In 
this  case  we  write 


lim 


fix)=cc. 


But  it  must  be  noted  that  this  equation  is  merely  symbolic, 
for  the  reason  that  the  symbol  oo  does  not  represeyit  a  num- 
ber. The  symbolic  equation  tells  us,  not  that  f(x)  ap- 
proaches some  vague,  indefinite,  very  large  limiting  value, 
but  that  it  increases  beyond  any  limit  whatever. 

Graphically  the  occurrence  of  such  a  discontinuity  means 
that  the  curve  y  =  f(x)  approaches  nearer  and  nearer  the 
line  x=  a^  usually  without  ever  reaching  it,  at  the  same 
time  receding  indefinitely  from  the  a:-axis. 

Examples:  (a)  As  x  approaches  0,  the  function 


1 


a;* 


becomes  infinite  (Fig.  2) 


linii  =00. 

a>>0  x^ 


Fig.  2 

(5)  The  function 


Fig.  3 


y  = 


X 


12  CALCULUS 

becomes  positively  or  negatively  infinite  according  as  x  ap- 
proaches 2  from  the  right  or  the  left  (Fig.  3) : 

lim       ^      =+Q0,  lira  _l_  =  _co. 

x-^'i^  a;  —  2  .r->2"' 2;  —  2 

14.  Function  with  infinite  argument.  We  have  fre- 
quently to  investigate  the  behavior  of  a  function  as  the 
argument  becomes  infinite. 

If  when  X  increases  indefinitely  the  difference  between 
f(x^  and  some  fixed  number  I  ultimately  becomes  and  re- 
mains numerically  less  than  any  preassigned  constant  how- 
ever small,  we  write 

Graphically  this  means  that  the  curve  y  =  f(x)  ap- 
proaches nearer  and  nearer  the  line  y  =  U  usually  without 
ever  reaching  it,  at  the  same  time  receding  indefinitely 
from  the  ?/-axis. 

Examples :    (^oC)    As   x  increases   indefinitely  in   either 

direction,  the  function  y  =  —^  approaches  0  (Fig.  2): 

lim  1  =  0. 

1  +  1 

(5)     lim  ^_+l_  lim  ^  =  1. 


1- 

x 


EXERCISES 


1.  Show  that  a  polynomial  is  continuous  for  all  values  of  x  (see 
Ex.  10,  p.  10). 

2.  For  what  values  of  a:  is  a  rational  fraction  discontinuous?- 

3.  For  what  values  of  x  is  the  function  discontinuous? 

a;2-4 

4.  Evaluate    lira  -^ — -  ■     Trace  the  curve  y  =  — — r« 
6.   Evaluate  («)      ^''\  ^+i ;    (b)     ^^"^   ^^^1. 

x->0+         x  x->0-       X 


FUNCTIONS.     LIMITS.     CONTINUITY  13 

6.   Evaluate    l^n  -r^^ll^. 


7. 

Evaluate 

4 

X  — 

(a) 

lim     3^2+  5x   . 
j-^ao  a;2  —  3  a;  —  1 

(&) 

lim             ^ 

a:->oo  3  x-  -  4 

(c) 

lim     10^ ; 

(^) 

Urn    10-; 

(e) 

lim   a;2  +  3z+l. 
x^^        a:  -  5       ' 

(/) 

lim  tana:. 

Ans.  (a)  3;  (c)  0;   (/)   non-existent. 

8.  Does  sin  x  approach  any  limit  as  x  becomes  infinite  ?     Does 
sin  Xq     Dogg  tan  x^ 

X  X 

9.  Show  that  as  x  approaches  0,  the  function  sin  -  oscillates  be- 

X 

tween  —  1  and  1,  without  approaching  any  limit. 

10.  Discuss  the  behavior  of  tan  -  near  the  origin. 

1 

11.  Discuss  the  behavior  of  10*  near  the  origin. 

12.  Evaluate    lim  2- sin-. 

13.  Is  the  function 

x-^  —  4 
continuous  at  x  =  2  ?     Can/(2)  be  so  defined  as  to  make/(a:)  contin- 
uous? 

14.  If  f(x)    is    continuous,   is    its    square    continuous?      Is   its 
reciprocal  ? 

15.  Given  two  continuous  functions,  what  can  be  said  of  the  con- 
tinuity of  theu'  sum  ?     Their  product?     Their  quotient? 

16.  Are  the  trigonometric  functions  continuous  for  all  values  of  the 
argument  ?    Discuss  fully. 


CHAPTER   II 


THE  DERIVATIVE 


15.  The  derivative.  We  return  now  to  the  problem 
(§6)  of  finding  the  rate  of  change  of  a  function,  or  the 
slope  of  a  curve. 

Given  a  function 

continuous  at  the  point  P  :  (x^  «/),  let  us  assign  to  x  an  ar- 
bitrary increment  A2;,  and  compute  the  corresponding  in- 
crement ^y  of  y.     We  have 

y  +  ^y^f(x  +  Lx), 

so  that 

^y=Ax  +  Lx)-f(x). 

Now  form  the  ratio 


Fig.  1 


dkX  Ax 


The  limit  of  the  ratio  —^  as  Ax  approaches  0  is  called  the 

Ax 

derivative  of  y  with  respect  to  x. 

The  derivative  is  designated  by  the  symbol  -^ : 

^=   lim   ^=    lim   fCx-hAx)-f(x) 

Other  commonly  used  symbols  for  the  derivative  are  y'^ 

The    operation    of    finding    the    derivative    is    called 
differentiation . 

14 


THE  DERIVATIVE 


15 


It  follows  from  §  6  that  the  derivative  of  a  function  is 
'  identical  with  its  rate  of  change.      Geometrically  the  deriva- 
tive of  a  function  is  the  slope  of  its  graph. 

Only  differ eyitiable  functions  (i.e.  those  having  a  deriva- 
tive) are  considered  in  this  hook.  In  some  cases  the  de- 
rivative may  fail  to  exist  for  particular  values  of  the 
argument,  but  such  values  are  either  excluded  or  subjected 
to  special  investigation. 

To  hnd  th6  derivative  of  a  given  function,  we  have 

merely  to  huild  up  the  "  difference-quotient "  — ^  and  then 

pass  to  the  limit  as  Ax  approaches  0.  It  will  be  remembered 
that  this  is  essentially  the  method  used  in  analytic  geometry 
to  find  the  slope  of  a  curve.  Since  Ax  and  Ay  approach  0 
together,  our  problem  is  to  find  the  limit  of  the  ratio  of  two 
infinitesimals  (cf.  §11).  In  general,  this  limit  cannot  be 
evaluated  until  some  suitable  transformation,  algebraic  or 

otherwise,  has  been  applied  to  the  quotient  _^  • 

Ax 

The  process  of  finding  the  derivative  is  illustrated  by 
the  following 

Examples :  (a)  Find   the   slope   of 
the  parabola 

y  =  2x^-6x-[-4: 

at  the  point  (ir,  3/) ;  at  the  point  (1,  0). 

If 

y=f(x)  =  2x^-6x  +  4:, 

then 
y -\- Ay  =  f(x -\-  Ax} 

=  2(x  -h  Axy-6(x  -h  Ax}  -h  4, 
A^  =  4  xAx  +  2  A?  —  6  Ax^ 

^  =  4.x+2Ax-6, 
Ax 

y'=   lini  ^  =  4  a;  _  6. 

^  Ax^O  Ax 


Fig.  4 


16  CALCULUS 

Hence  the  slope  at  any  point  (x,  y)  is  4  a^  —  6 ;  in  par- 
ticular, the  slope  at  the  point  (1,  0)  is  —  2. 


(5)  Given 

t            dt 

We  have 

1 

*  +  ^'-«  +  A«' 

A„_     1          1 

«  +  A«      t 

Reducing  the  fractions  in  the  right  member  to  a  common 
denominator,  we  find 

^^  ^  ^  -  (^  +  AQ  ^     -A/^ 

Whence 

Ag^     -i 

A^  "  (^  +  AO*' 
^=  lim  4^=_i. 

Geometrically  this  means  that  the  slope  of  the  hyper- 
bola s  =  -  at  the  point  (^,  s)  is  — -• 

Zl  Li 

(c)  Find  the  rate   of  change  of  the  function  y  =  V^ 
at  the  point  (rr,  y);   at  the  point  (4,  2). 

If 

y=^x, 

then 

y  -\-  l^y  =  Va:  +  Aa;, 

A?/  =  Vo:;  H-  Aa;  —  Va: 


=  ( Va;  4-  b^x  —  Va;) 

_    (a:  +  Aa;)  —  a; 

Va;  +  Aa:;  4-  Va; 
A^ ■ 

Va;  +  Aa:  +  Va; 


Va:  +  Aa:  4-  Va; 

Va:  +  Aa;  +  Va; 


THE   DERIVATIVE  17 


^=   liiii    %^     1     . 
At  the  point  (4,  2),  the  rate  of  change  is 


EXERCISES 


^X    1^=4  ■! 


Find  the  slopes  of  the  following  curves  at  the  points  indicated. 

1.  y  =  X  —  x^  2ht  (x,  y) ;  at  :c  =  2.     Trace  the  curve. 

2.  3/  =  a:^  +  1  at  (z,  ?/).     Trace  the  curve. 

3.  y  =  x^  —  x^  at  the  points  where  the  curve  crosses  the  a:-axis. 
Trace  the  curve. 


4.    3/  = at  X  =  2.  '  Ans. 

X  -\-\ 

6.   y  =  —  at  a:  =  2. 

x^ 

6.  y  =  a:^  -  3  a;2  +  2  at  (a;,  3/). 

7.  3/  =  X  +  -  at  a;  =  2 . 

8.  If  3/  =  -,  find  '^. 

a:^  6/a; 


9.   li  y  =  V3  —  a:,  find  y' .  Ans. 


-  1 


2\/3-x 


10.    If/(:c)= ^^ -,find/'(a:).  Ans.  ^ 


(1  -  xy  (1-  xy 

11.    If  s  is  measured  in  feet  and  t  in  seconds,  find  the  rate  at  which 
s  is  changing  at  the  end  of  2  seconds  when 

(a)  s  =- ^;     (^)  s  =Vt  -\-  1.      Ans.    (a)  |  ft.  per  second. 

X  1 

. .    12.    At  what  points  does  the  curve  y  = have  the  slope  -? 

a:  +  1  4 

Am.    (1,  i),  (-3,1). 

13.    Differentiate  3/ = — -•  Ans.  ^^-■. 

Va;  2xi 

^14.   Find  ^  if  r  =  $1  Ans.    ^oK 

du 

c 


18  CALCULUS 

15.    Differentiate  y  = 

^       (x-l)2 


16.   If  f{x)  =  Vrt2  _  a;2^  find  f'(x).  Ans. 


—  X 


Va2 


17.    Find  the    angle   between  the    curve  y  = •  and  the  line 

a;  +  1 

y  =  X  at  each  point  of  intersection. 

16.  Higher  derivatives.  The  derivative  of  ^  with  re- 
spect to  X  is  itself  a  function  of  x.  The  derivative  of  the 
first  derivative  is  called  the  second  derivative^  and  is  written 

— ^  (read  d  second  7/  over  dx  square);  the  derivative  of 

(JiX 

the  second  derivative  is  called  the  third  derivative,  written 
^-  etc 

Other  symbols  for  the  higher  derivatives  are  y",  i/'^',  • .  •  ; 
D.%D.%...;f"(x-),f"'Cx%.... 

Example:     In  example  (a),  §  15,  we  found    . 

y'  =  4:X  —  6. 

Hence 

■  y  +  A^'  =  4:(x  -h  Ax^  —  6, 

Ay'  =  4  Aa;, 

Ax 
y"  =  lim  — ^  =  4. 

Aa:->0  Aa^ 

In  this  case  all  the  higher  derivatives  are  0. 

EXERCISES 

1.  Find  y"  and  y'"  in  Exs.  2,  3,  5,  p.  17. 

2.  In  example  (?>),  §  15,  find 

3.  In  Ex.  10,  p.  17,  find/"(x). 

4.  In  Ex.  11,  p.  17,  find  how  fast  -y-  is  changing  when  t  =  2  seconds. 


CHAPTER   III 
DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS 

17.  Introduction.  In  this  and  a  later  chapter  (V)  we 
develop  certain  standard  formulas  by  means  of  which  any 
elementary  function  may  be  differentiated.  The  use  of 
these  formulas  effects  a  great  saving  of  time,  and  obviates 
the  necessity  of  evaluating  a  special  limit  in  every  problem. 

The  formulas  of  §§  19-20  are  direct  consequences  of  the 
definition  of  the  derivative,  and  are  valid  for  all  functions 
{i.e.  all  functions  that  are  continuous.,  one-valued.^  and 
differentiahle ;  see  §§  12,  5,  15). 

18.  Derivative  of  a  constant.  We  note  first  that  the 
derivative  of  a  constant  is  0: 

(1)  ^=0. 

^  ^  dx 

For,  if  y  =  c,  then  no  matter  what  the  values  of  x  and 
Aa:  may  be,  y  will  remain  unchanged,  and  hence  Ay  =  0  : 

^  =  0,^=   lim  ^=0. 

Ax  dx       Ax->o  Aa; 

The  line  ?/  =  <?  is  parallel  to  OX  ;  its  slope  is  everywhere  0. 

19.  Derivative  of  a  sum  ;  a  product ;  a  quotient.  If  u  and 
V  are  functions  of  ic,  the  following  formulas  are  true  by  the 
definition  of  the  derivative  : 

•ON  d  .     ,     N      du  ,  dv 

dx  dx     dx 

(3) 

(4) 

axw/ 

19 


l(..)= 

dv  ,     du 
dx        dx 

dx\vJ 

du        dv 
V —  —  u  — 

^dx       dx 

v" 

20  CALCULUS 

These  formulas  may  be  stated  in  words  as  follows  : 

(2)  The  derivative  of  the  sum  of  two  functions  is  equal  to 
the  sum  of  their  derivatives. 

(3)  The  derivative  of  the  product  of  two  functions  is  equal 
to  the  first  function  times  the  derivative  of  the  second  plus  the 
second  times  the  derivative  of  the  first. 

(4)  The  derivative  of  the  quotient  of  two  functions  is  equal 
to  the  denominator  times  the  derivative  of  the  numerator 
minus  the  numerator  times  the  derivative  of  the  denominator., 
divided  hy  the  square  of  the  denominator. 

Proof  of  (2) ;  Let  x  assume  an  increment  A:r,  and  de- 
note by  ^u  and  Av  the  corresponding  increments  of  u  and  v. 
Then 

y  =  U+  V^ 
y  +  Ay  =  u-\-  ^u  +  V  -h  A?;, 
Ay  =  Ai^  +  Av, 
Ay  _  Aifc      A?; 
Aa;      ^x      A2; 
^y  —    y      ^y  _d'^  ,  dv 
dx      Ax->o  Aa;      dx      dx 
Proof  of  (3): 

y  =  uv, 
y  -\-  Ay  =  (^u  +  Aw)(v  +  A?;), 
Ay  =  uAv  H-  vAii  +  AuAv^ 
Ay        Av  ,     Au  ,    .     Av 
Ax        Ax        Ax  Ax 


^^  lim   ^  =  u—-{-v~ 
dx      Ax->oA2;         dx        dx 


Proof  of  C^): 


u 

y  =  -'' 

V 


.         u-\-  Au 

y  +  ^y=    T.  ^ 

V  +  Av 

.         u-\-  Au      u      uv  +  vAu  —  uv  —  uAv 
Ay  =  • -— =  — — — -— , 

V  -\-  Av      V  {y  -\-  Av)v 


DIFFERENTIATION  OF  ALGEBRAIC    FUNCTIONS    21 

Au        Av 
V u  — 

A^        Ax        Ax 

Ax~   Qv  +  Av)v 

du        dv 
V- u  — 

dg  _  i[^^  A^  _     dx        dx 
dx~  A-^-^oAa:  v^ 

Formulas  (2)  and  (3)  can  be  extended  to  the  case  where 
n    functions   are    involved.      For    three    functions,    (3) 

becomes 

d                  du  ,        dv  ,        dw 
—uvw  =  vw—~  +WU \-  uv 

dx  dx  dx  dx 

In  the  special  case  when  t*  =  (?,  a  constant,  (3)  and  (4) 
become 

(3')  —cv=c--. 

dx  dx 

dv 

r4'^  -lf-__^ 

^    ^  dxv~       v''  ' 

20.  Derivative  of  a  function  of  a  function.  A  function 
is  sometimes  expressed  in  terms  of  an  auxiliary  variable 
which  in  turn  is  a  function  of  the  independent  variable ; 
for  example, 

?/  =  5  w^  +  2  i*,  where  u  =  x^  -\-  ox-{-l. 
The  variable  u  in    the   first  equation   may   of   course   be 

replaced  by  its  value  in  terms  of  a:,  and  — ^    can  then  be 

dx 

determined  directly ;  but  it  is  desirable  to  have  a  formula 
by  which  — ^  can  be  found  without  eliminating  u. 

dX 

Let  y  =f(u)^  where  u  =  4>(_x). 

Assign  to  X  an  increment  Ax,  and  denote  by  Au  and  Ai/ 
the  corresponding  changes  in  u  and  ?/.     Then 

A?/  _  A?/     Au 

Ax      Au     Ax 

and,  passing  to  the  limit,  we  find 


22  CALCULUS 

T      Ay       T       Ay     -,.      Au 
lim  — ^  =  lim  — ^  •   lim  — , 

Ax->^Ax      aj:->o  A-w    Ax->oAa; 
or 

^  ^  dx     du    dx 

This  very  important  formula  is  easily  remembered  from 
the  fact  that  inform  it  is  a  mere  identity. 

21.    Derivative  of  jc**,  n  a  positive  integer.     If 

?/  =  a;", 
where  w  is  a  positive- integer,  then 

(1)  4^  1=^^^'^" 

For, 
y  +Ay=(x  +  Axy 

=  3;"H-na;"-^A2;+  ^^^^^  ~  ^^ x^'-'^Ax  +  •••  +  A^", 

2  I 

Ay  =  na^"~iAa;  +  ^^^7 — ^^""^A^;^  +  •  •  •  +  Ax"", 

2 ! 

^  =  nx'^-^  H-  ^(^-  ^)a:^-2Aa;  +  -.  +  A^"-^ 

Ax  2! 

— ^  =   lim  — ^  =  na;'^"^ 
6?a:       A^->.o  Ax 

In  particular,  if  ?i  =  1,  i.e.  ii  y  =  x.,     . 

dx  __  H 

dx 

which  is  obvious  geometrically. 

By  means  of  (4'),  it  can  be  shown  that  (1)  is  true  when 

w  is  a  negative  integer. 

Examples:  (a)  Find  the  derivative  of 

y  =  ?>7^  +  l  x^+1. 

dx  dx  dx 


=  9  a:2  +  14  a:. 
(5)  Differentiate       y  = 


x^ 


a;-f  3 


I 


DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS    23 


y 


^  ax  ax 

^  {x+-6f 

_{x  +  ^')2  X  -  x^  ^  x^  +  Q  X 
(2; +3)2  ix+2>y 


EXERCISES 

Differentiate  the  following  functions. 

1.  (a)  y  =  bx^-2x;  (b)  s  =  t  -'St^  +  t^. 

2.  (a)  ?/  =  x4  -  3  x3  -  2  a:2  _  1 ;  (6)  y  ^  x\o  x^  +  3). 
,.3.    (a)  y  =  l-2x-3a;5;   (6)  y  ={x^  -  l)(x2  +  3x  +  2). 

1  +a:2 


^      a;2-l  (x2-l)2 

-   5.    y  =  L±2^LZL^.  6.    2/=(xB-l)(:.3+i). 

x^  +  6 

,7.    v  =  ^^-  ■  8.   y  =  (l+^)a-2x). 

1  -  a:  X 

9.    If  ^^=-^'-^,   find^.  Am.    -\ 

X  dx^  x' 

10.  Find  —  (5  a:3  +  7  a:2  +  8  a:)(x2  +  3  x  +  4). 

dx 

11.  Differentiate  ?/  =(a:  +  l)(a:  +  2)(x  +  3). 

12.  If  a:  =  -,  find  —  • 

t^  dfi 


_  /v^+i 


(1-0 


13.  If  F{t)  =  {-!—Y,  find  F'(0.  ^m. 

9      3 

14.  Find  the  rate  of  change  of     s  =  t  —-  -\ 

t      t^ 

15.  In  the  proof  of  (1),  §  21,  why  is  n  assumed  to  be  a  positive 
integer? 

16.  Show  directly  from  the  definition  of  the  derivative  that  for- 
mula (1)  of  §  21  holds  when  n  =  ±  I. 

17.  Find  ^  if  ^/  =  2  w^  _  4,  m  =  3a;2  +  1. 

dx 

18.  Find  the  slope  of  the  curve  y  =  x (x  +  l)(x  +  2)  at  the  points 
where  it  crosses  the  x-axis.     Trace  the  curve. 


24  CALCULUS 

19.  At  what  points  is  the  tangent  to  the  curve  y  =:(x  —  3) '-(a:  —  2) 
parallel  to  0X1     Trace  the  curve. 

20.  Prove  formula  (l)of  §  21  when  n  is  a  negative  integer. 

21.  Given  a  polynomial  of  the  n-th  degree,  prove  that  all  the  deriva- 
tives after  the  n-th  are  identically  0. 

22.  Derivative  of  x^,  n  fractionaL  By  means  of  formula 
(5),  the  power  formula  (1)  of  §  21  can  be  extended  at  once 
to  the  case  when  n  is  a  rational  fraction. 

If  2 

where  p  and  q  are  positive  integers,  then 

y^  =  xP. 
Differentiating  each  member  of  this  equation  with  respect 
to  x^  we  find,  by  (5), 

dx      q  y'^~^      q    y'^ 

q  x^  q 

This  shows  that  formula  (1)  of  §  21  holds  even  when  n  is 
a  positive  rational  fraction. 

By  using  (4'),  the  formula  can  be  shown  to  hold  when 
?i  is  a  negative  rational  fraction. 

In  more  advanced  texts  it  is  proved  that  the  power 
formula  holds  when  n  is  irrational,  and  hence  is  valid  for 
all  values  of  n. 

23.  The  general  power  formula.     Suppose 
y  =  u^^  where  u  =  </)(^). 


Then 


du 


and  we  have  by  (5) 

(6)  ^U^  =  nu^-^~ 

^  dx  dx 


DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS    25 


An  important  special  case  of  this  formula  is  the  case 


du 

^«/x  d     .—       dx 

(6')  ^^u  =  — -=.' 

Example :     Find  the  derivative  of 

This  function  is  of  the  form  w",  with  it  =  3  a;^  -)-  1,  ?i  =  4. 
Hence  (6)  gives 

^  =  4(3  2;2  +  1)3 .  6  re  =  24  a;(3  x^  +  1)3. 
ax 

EXERCISES 

Differentiate  the  following  functions. 
1.    2/ =z  a;3  +  3a;2.  2.   y  =  — 


X2  +  1 

-  3.   y=(2x+iy.  Ans.   /•  =  10(2  x  +  l)^. 

4.    ?/  =  (a;3  +  5  a;2  +  7)2.  ^n.-?.   y'  =  (Qx^  +  20  x)  (x^  +  5x^  +  7). 

.  b.   y  —  Vl  —  "d  x^.  Q.   y  =  — — — ^ . 

Va:^  —  a: 

7.    ?/ =  fj  -ix"^  +  3  a:.  8.    y  =(x^  -  5xy(8x-7y. 

-  (3a;«-7a;+  1)3  ..  .  „   ,     *^+ 

(1  —  x)^ 


"11.   ?/=(3x  +  2)3  +  (5x  +  7)2.         12.   Z/=^^ 


.+  2x 


2x 
13.   y  =  ^^^^^.  14.   ^/  =  (1  -  x2)(l  +  x2)2. 

>.  15.    ?/ = —  — •  vlns.    ?/' = 


V(a2_x2)3  -        (a2-x2)l 

16.   Find  the  slope  of  the  hyperbola  x'^  —  y^  =  12  at  (4,  —  2). 

Ans.  -2. 


*47.   If  <^(y)  =^       ^      ,  find  <^'(tO,  <^"(0.  <^"'(^') 

8(1  -t;)2 


18.  Find  1  Vl  -  s  •  Vl  +  s^. 

19.  If  y  =2  V^,  find  v".  20.    Find  —  (at  -  xl)i 


26  CALCULUS 

21.  Differentiate  ^  =(  )  .  Ans.   y' = ^ 

\^1  +  Vl  -  x'^J  x-\/l  -  a;2 

22.  Find  the  slope  of  the  curve  y  =  (x^  —  l)^  at  each  of  the  points 
where  y"  =  0.     Trace  the  curve. 

23.  Draw  the  graph  of  the  function  y  =  a:"  for  n  =  i,  1,  |,  2,  3. 

24.  Implicit  functions.  Up  to  this  point  we  have  been 
concerned  with  functions  defined  explicitly  by  an  equation 
of  the  form  i/=f(x}. 

It  may  happen,  however,  that  x  and  i/  are  connected  by  an 
equation  not  solved  for  i/  ;  for  example, 

x^  -^  y^  =.  aP'. 

In  such  a  case  y  is  called  an  implicit  function  of  x^  and 
the  relation  is  expressed  by  writing 

The  definition  becomes  explicit  if  we  solve  for  y ;  in  the 

above  example,  

y  =  ±  Va^  —  x'^. 

25.  Differentiation  of  implicit  functions.  To  find  the 
derivative  of  an  implicit  function,  we  proceed  as  follows  : 

Differentiate  each  term  of  the  equation 

Fix,  y)  =  0, 
hearing  in  mind  that^  owing  to  the  equation^  y  is  a  fu7iction 
of  X. 

Example  :     Find  the  slope  of  the  ellipse 
x^  —  xy  +  y"^  +  x^l 
at  the  point  (a:,  ?/). 
We  have  d 


{x^  —  xyi-y^-^-x^^Oy 
ax 


or  by  (6)  and  (5) 


2x-x^-y-h2y^  +  l  =  0, 
dx  dx 

dy  _  2x  —  y  -\-l 

dx  X  —  2y 


DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS    27 


26.    Inverse  functions.     The  equation 

(1)  y=/(^) 

defines  x  implicitly  as  a  function  of  y ;  when  solved  for  ^, 
it  takes  the  form 

(2)  ^  =  <^(y> 

The  function  (^(^)  is  called  the  inverse  of  the  direct  func- 
tion f{x). 

For  example, 

(a)   if  ^  =  ^^       then  x=^±.  Vy  ; 


(J)  \i  y  =  ±  Va^  —  a:^,     then  a:  =  ±  Va^ 

(c)  if  ?/  =  «•%  then  x  =  log^-grT""' 
In  each  case  the  second  func- 
tion is  the  inverse  of  the  first. 

To  construct  the  graph  of 
the  inverse  function  from 
that  of  the  direct  function, 
we  have  only  to  interchange  x 
and  y^  which  amounts  to  a 
reflection  in  the  line  y  =  x. 
This  is  shown  in  the  figure 
for  example  (a)  above. 

If  f(x)  and  (f>(^y)  are  in- 
verse functions,  then 

1 


y 


(3) 

For,  since 


6.x 


J_ 
Ay 


Fig.  5 


(*'(2/)^0). 


we  have,  passing  to  the  limit, 

dy^l_ 
dx     dx'' 
dy 
which  by  (1)  and  (2)  is  the  desired  formula. 


28  CALCULUS 


EXERCISES 

1.  Express  y  explicitly  as  a  function  of  x,  when 

(a)  x^  —  y^  —  1\  (6)  2  a;?/  +  2/2  —  4  I 

(c)   sin  (x  +  2/)  =  1  ;  {d)  xa^  =  1. 

Find  the  slopes  of  the  following  curves  at  the  points  indicated. 

2.  a;2  +  7/2  =  25  at  (x,  y)  ;  at  (3,  4).  Ans,  -  -;    -  -• 

y       4 

-  3.    a:2  +  x?/  +  ?/2  =  3  at  (1,  1).  ^m.   -  1. 
4.    2  a:2  +  2  2/^  -  9  a:?/  =  0  at  (1,  2). 

-  5.   xy^  =  3  at  (3,  1).     Do  this  in  two  ways. 

Find  ^  in  the  following  cases. 

dx  ^ 

6. 1-^^=1.  Ans.    — 

a^      h^  a^y 

7.  {x  —  y)  (x-  +  yy  =  a^.  8.    x^  -\-  y^  —  3  axy  =  0. 

,9.  x-\-  V(2x  -  'dyy  =  0.  10.    x^y^  +  2  x^y  -  xy^  +  2  =  0. 

11.  a;  =  (1  —  3  ?/)2.     Solve  in  two  ways. 

12.  If  i/2  =  4  ax,  find  3/".     Cf.  Ex.  19,  p.  25.  Ans.  -  ^. 

yZ 

~"  13.   Prove  that  a  tangent  to  a  circle  is  perpendicular  to  the  radius 
drawn  to  the  point  of  contact. 

14.  If  a:2  +  2/^  =  a",  find  y" . 

15.  Show  that 

d'^y 

d^x  dx'^  f  dv         \ 


df 


\dxl 


16.  Find  the  inverse  of  each  of  the  following  functions : 

(a)?/  =  3a:-4;  (&)2/  =  £z^; 

X  —  2 

(c)  y  =  (x2-l)2;  (f7)  2/=logioa:. 

17.  In  Ex.  16,  are  the  direct  functions  one-valued?  Are  the  in- 
verse functions? 

18.  In  Ex.  16  (a),  (c),  verify  formula  (3),  §26. 

19.  In  Ex.  16  (a),  (c),  verify  that  the  graph  of  the  inverse  function 
may  be  found  from  that  of  the  direct  function  by  reflection  in  the  line 
y  =  x. 

20.  In  Ex.  16  (&),  (c),  discuss  the  continuity  of  the  direct  and  in- 
verse functions. 


CHAPTER  IV 


GEOMETRIC  APPLICATIONS 


27.    Tangents  and  normals  to  curves.     It  is  known  from 
analytic  geometry  that  the  equation  of  a  line  through  the 
point  (a^Q,  y^)  with  slope  m  i% 
(1)  y-y^  =  m(ix-XQ). 

Let  P  :  (a-Q,  y^)  be  a  point  on  the  curve 
F(x.  y-)  =  0, 

and  denote  by  ^q'  the  value  of  the  derivative  at  the  point 
(a^Q,  ?/q).  The  equation  of  the  tangent  to  the  curve  at  P 
c^  then  be  written,  by  (1),  in  the  form 

The  equation  of  the  normal  —  i.e.  the  line  perpendicular 
to  the  tangent  at  the  point  of  contact  —  can  be  found  at 
once  from  that  of  the  tangent,  by  recalling  that  if  two 
lines  are  perpendicular,  the  slope  of  one  is  the  negative 
reciprocal  of  the  slope  of  the  other. 

Examples:  (a)  Find  the  tangent  and  normal  to  the 
ellipse 

42:2  +  9^/2  =  40 

at  the  point  (1,  —  2). 
,  t      We  have 

Sx-hlSyy'=0, 
hence 


'^  =  2/0  =  - 


4:X' 


(1,  -2) 


Therefore  the   equation  of^the 
tangent  is 

2/  +  2=f(a:-l), 
that  of  the  normal  is 


Fig.  6 


y  +  2  =  -ICx-l). 
29 


30 


CALCULUS 


Fig.  7 


(6)   Find  the  equation  of  a  tangent  to  the  curve  y  =  7^ 
parallel  to  the  line  «/  =  3  :r  +  1. 

The  slope  of  the  required  tangent  is  3.     But  the  slope 
at  any  point  (a;,  ?/)  of  the  curve  y  =  7?\'& 
y=3a:2. 

Hence  the  coordinates  of  the  point- of 
contact  are  found  by  solving  the  simul- 
taneous equations 

3  a;^  =  3,     y  =^Qi^. 

This  gives  the  points  (1, 1),  (—  1,  —  1), 
and  the  required  tangents  are 
^  _  1  =  3(^  _  1),      ^  +  1  =  3(a;  4- 1). 
28.    Length  of   tangent,    subtangent,    normal,    and   sub- 
normal.     Let  P :  {x,  y}  be  a  point  on  the  curve 

FCx,  2/)  =  0. 
The  segment  TP  of  the  tangent  intercepted  between  the 
point  of  tangency  and  the  2:-axis  is  called  the  length  of  the 
tangent;  its  projection   TQ  on  OX  is 
called    the    length    of  the   subtangent. 
The  segment  iVP  of  the  normal  inter- 
cepted between  P  and   the  a:-axis  is 
called  the  length  of  the  normal;  its  pro- 
jection $iV  on  the  a:-axis  is  called  the 
length  of  the  subnormal. 

It  is  customary  to  consider  all  these 
lengths  as  essentially  positive.     They  are   evidently  de- 
termined by  the  coordinates  (a;,  y}  of  the  point  P  and  the 

slope  at  P. 

EXERCISES 

Find  the  tangent  and  normal  to  each  of  the  following  curves  at  the 


points  indicated. 

1.    (a)  y  =  1  —  X  —  x^  at  (1,  —  1)  ; 

(c)   x2  +  y2  =  25at(-3,  4); 


(b)   xy  =  2  a-2  at  (a,  2  a)  ; 
{d)  y  =  a:  +  -  at  (1,  2). 


Ans.    (a)  ^  +  1  =  -  3(a;  -  1),  i/  +  1  =  \{x  -  1). 


GEOMETRIC   APPLICATIONS  31 

"^  x^  —  2  xy  -\-  2  y^  —  X  =  0  at  the  points  where  x  =  1. 

Ans.    At  (1,  0); 
2y  =  x-l,  i/-\-2x  =  2;   at  (1,  l):2?/  =  a:  +  l,  ?/  +  2a:=3. 

"St   y  =  — — -  at  x  =  2a.  Ans.    x-\-2u  =  ia,  y  =  2x  —  ^a. 

4.  Find  the  equation  of  the  tangent  to 

(a)  y^  =  iax  at  (x^,  y^)  ;   (&)    ^^  ±  |^  zz:  1  at  (xq,  y^) . 

Ans.   (a)  y,y  =  2a(x  +  x,);   (b)    ^  ±  M  ^  i. 

5.  Find  the  subtangent,  subnormal,  tangent,  and  normal  lengths 
in  each  of  the  cases  of  Ex.  1.     Draw  a  figure  in  each  case. 

6.  Find  the  angle  between  the  parabolas  y^  =  x,  y  =  x^  at  each  of 
their  points  of  intersection. 

7.  Find  the  tangent  and  normal  to  the  curve  y"^  =  2  x^  —  x^  at  the 
points  X  =  1.  Ans.    At  (1,  1)  : 

^     2y  =  x-{-l,y  +  2x=3',  at  (1,  -  1)  :  x  +  2y  +  1  =  0,  y  =  2x  -  3. 

_  8.  Show  that  the  subtangent  to  the  parabola  y^  =  4:  ax  is  bisected 
at  the  vertex,  and  that  the  subnormal  is  constant.  Hence  give  a 
geometric  construction  for  drawing  the  tangent  and  normal;  also 
show  how  to  find  the  focus  of  a  parabola  if  the  axis  is  given. 

9.   Find  a  tangent  to  the  parabola  y^  =  4ax  making  an  angle  of 
45"^  with  the  a:-axis.  Ans.  y  =  x  +  a. 

10.  Find  the  tangents  to  the  hyperbola  '^x^  —  9y-  -{■  36  =  0  per- 
pendicular to  the  line  2  y  ■\-  5x  =  10.  Ans.   2x— 5y=±8. 

—11.    Find  a  tangent  to  the  curve  y  =  1  —  x^   parallel   to  the  line 
X  —  2  y  =  d. 

12.  Find  a  normal  to  the  parabola  y  =  x'^  perpendicular  to  the 
line  3  X  —  2  y  =  1. 

-13.   Show   that   the   portion   of   the  tangent   to   the  hypocycloid 

2  2  2 

x3^y'5  =  «3  intercepted  between  the  axes  is  constant. 

14.  Find  the  tangents  to  the  circle  x^  +  y'^  =  5  which  are  parallel 
to  the  line  x  -r  3  y  =  0.     Draw  the  figure. 

Ans.   y±iV2  =  -l(x±iV2). 

15.  Find  the  tangents  to  the  curve 

y  =  \  x^  —  x'^  -{-  5  X 
which  make  an  angle  of  45°  with  the  x-axis.     Plot  the  curve. 

16.  Find  the  angle  between  the  line  y  =  —  2  x  and  the  curve 
y  =  x2(l  —  x)  at  each  point  of  intersection. 


32  CALCULUS 

17.  Find  the  equation  of  a  tangent  to  the  curve  y^  =  1  —  x  parallel 
to  the  ?/-axis.     Trace  the  curve. 

^18.    Show  that  the  area  of  the  triangle  formed  by  the  coordinate 
axes  and  the  tangent  to  the  hyperbola  2  xij  =  a^  is  constant. 

19.  Show  that  the  length  of  the  normal  is  constant  (equal  to  a)  in 

the  circle 

(x  —  c)^  +  7j'^  =  a^, 

where  c  is  arbitrary,  and  explain  geometrically. 

20.  Show  that  the  sum  of  the  intercepts  on  the  axes  of  the  tangent 

1         1         1  . 
to  the  parabola  x^  +  y^  =  a^  is  constant. 

21.  Show  that,  in  the  curve  y"^  =  ax,  the  subtangent  is  n  times  the 
abscissa  of  the  point  of  contact.  Hence  show  how  to  draw  the  tan- 
gent at  any  point  of  the  curve  y  —  ax". 

22.  Find  the  length  of  the  tangent,  subtangent,  normal,  and  sub- 
normal to  the  curve  y  =  f(x)  at  the  point  {x,  ?/).  Ans.    Tangent, 

■^  Vl  +  y''^\,   subtangent,  ^;    normal,  ^/Vl  +  y''^;    subnormal,  yy' . 

y'  y' 

29.  Increasing  and  decreasing  functions.  In  studying 
the  properties  of  a  function 

it  is  usually  of  great  assistance  to  represent  the  function 
graphically.  In  tracing  a  curve,  it  is  well  to  begin  by 
locating  several  points,  e.g.  the  intersections  with  the 
axes,  and  finding  the  slope  at  those  points ;  it  is  also 
useful  to  note  the  behavior  of  y  for  large  positive  and 
negative  values  of  x. 

In  addition  to  giving  the  slope  at  any  point,  the  differ- 
ential calculus  is  of  assistance  in  a  variety  of  other  ways, 
as  will  be  shown  in  the  next  few  articles. 

We  shall  assume  as  usual  that  the  function  in  question  is 
one-valued,  continuous,  and  differentiable. 

We  note  first  that,  as  x  increases,  the  curve  rises  if  the 
slope  is  positive,  as  on  the  arc  AB  (Fig.  9);  it  falls  if  the 
slope  is  negative.,  as  along  BD : 

li  y'  >  0,  ?/  increases  ; 

If  y^  <  0.,  y  decreases. 


GEOMETRIC   APPLICATIONS 


33 


At  a  point  such  as  B  (Fig.  9), 
y 

F 


Fig.  9 


Of  course  this  also  appears  at  once  from  the  fact  that  y' 
is  the  rate  of  change  of  y. 

30.  Maxima  and  minima. 

where  the  function  is  al- 
gebraically greater  than 
at  any  neighboring  point, 
the  function  is  said  to 
have  a  maximum^  value ^ 
and  the  point  is  called  a 
maximum  point.  Simi- 
larly,  at  a  point  such  as  D 
the  function  has  a  minimum  value.  It  is  evident  that  at 
such  a  point  the  tangent  is  parallel  to  OX ;  i.e. 

/  =  0. 

But  the  vanishing  of  the  derivative  does  not  mean  that 
the  function  is  necessarily  a  maximum  or  a  minimum  ; 
the  tangent  is  parallel  to  OX  at  ^,  yet  the  function  is 
neither  a  maximum  nor  a  minimum  there.  It  appears 
from  the  figure  that  the  test  is  as  follows  : 

At  a  point  where  ?/'  =  0,  if  y'  changes  from  positive 
to  negative  (as  x  increases),  y  is  a  maximum ;  if  y' 
changes  from  negative  to  positive,  y  is  a  minimum  ;  If 
y'  does  not  change  sign,  y  is  neither  a  maximum  nor  a 
minimum. 

Since  the  function  is  continuous,  the  maxima  and  minima 
must  alternate :  between  two  maxima  there  must  be  a 
minimum,  and  vice  versa. 

The  points  at  which  «/'  =  0  are  called  critical  points, 
and  the  corresponding  values  of  x  are  the  critical  values 
of  X. 

31.  Concavity.  The  second  derivative  is  the  rate  of 
change  of  the  first  derivative.  It  follows  that  wKen  ^  is 
positive,  y  is  increasing :  as  x  increases  the  tangent  turns 
in  counterclockwise  sense  and  the  curve  is  concave  upward. 


34 


CALCULUS 


When  «/''   is  negative,  y^  decreases :   the^curve  is  concave 
downward. 

At  a  maximum  point  the  curve  is  concave  downward,  and 
hence  «/'^,  if  it  is  not  0,  must  be  negative.  At  a  minimum 
«/",  if  not  0,  must  be  positive.  If  the  second  derivative  is 
easily  obtained  and  if  it  does  not  happen  to  be  0  at  the 
critical  point  in  question,  it  is  usually  more  convenient  to 
determine  whether  we  have  a  maximum  or  a  minimum  by 
finding  the  sign  of  «/'' ;  but  the  test  of  §  30  has  the  ad- 
vantage of  being  perfectly  general.  However,  in  practice 
other  considerations  usually  enable  us  to  distinguish  be- 
tween maxima  and  minima  without  the  application  of  either 
of  these  tests. 

Example:  Find  the  maximum  and  minimum  values  of 
the  function 

y  =  x^  —  '^  X., 
and  trace  the  curve. 

This  curve  crosses  the  a:-axis  at  a:  =  0,  ±  VS.     Since 

^'  =  3  2:2  -  3, 

the  slope  at  (0,  0)  is  —  3,  at  ( ±  V3,  0)  it  is  6. 
Setting 

/  =  0, 
we  find  the  critical  points  (—1,  2),  (1,  —2).     When  x 
is  large  and  negative,  y  is  large  and  negative ;   when  x  is 
large  and  positive,  y  is  large  and  positive.     It  is  therefore 

clear  that  the  curve  must  rise  to  a 
maximum  at  (  —  1,  2),  fall  to  a  mini- 
mum at  (1,  —  2),  and  then  rise  in- 
definitely. These  conclusions  may  be 
verified  as  follows.  The  second  de- 
rivative, 

y'^  =  6  a:, 

is  negative  at   a;  =  —  1,   so  that \  the 
Fig.  10  point  ( —  1,  2)  is  a  maximum  ;  y\  is 


GEOMETRIC   APPLICATIONS  35 

positive    at   x=l,   hence    (1,  —  2),  is  a  minimum.     The 
curve  is  shown  in  the  figure. 

EXERCISES 

Examine   the  following  functions  for  maxima  and  minima,  and 
trace  the  curves. 

1.   y  =  x{x  +  5).  2.    y  =  x^  -  2x'^  +  x. 

■^3.   y  =  x^x'^  -  8).  4.    ?/  =  (x2  -  4)2. 

..  b.   y  ■=  x^  +1.  6.    ?/  =  a;3. 


9.    ^  =  ^,-  -r    10.    a:  =0  +  1)3. 

32.  Points  of  inflection.  A  point  at  which  the  curve 
changes  from  concave  upward  to  concave  downward,  or 
vice  versa,  is  called  a  point  of  inflection.  At  a  point  of 
inflection  the  tangent  reverses  the  sense  in  which  it  turns, 
so  that  y"  changes  sign.  Hence  at  such  a  point  y" ^  if  it 
is  continuous,  must  vanish.  In  Fig.  9  the  points  (7,  U,  F 
are  points  of  inflection. 

Since  y^' — i.e.  the  rate  of  change  of  the  slope  — 
vanishes  at  a  point  of  inflection,  the  tangent  is  sometimes 
said  to  be  stationary  for  an  instant  at  such  a  point,  and  in 
the  neighborhood  of  the  point  it  turns  very  slowly. 
Hence  the  inflectional  tangent  agrees  more  closely  with 
the  curve  near  its  point  of  contact  than  does  an  ordinary 
tangent ;  it  is  therefore  especially  useful  in  tracing  the 
curve  to  draw  the  tangent  at  each  point  of  inflection. 

33.  Summary  of  tests  for  maxima  and  minima,  etc. 
The  results  of  §§  29-32  may  be  summarized  as  follows : 

Let  the  function  y  =/  (x)  and  its  first  and  second 
derivatives  be  one- valued  and  continuous. 

(a)  In  an  interval  where  y'  >  0,  the  curve  rises;  where 
y'  <  0,  the  curve  falls. 

(J)  A  point  where  y'  =  0  is  a  maximum  or  a  minimum 
point.,  unless  at  the  same  point  y"  =  0,  in  ivhich  case  see  (e) 


36  CALCULUS 

below.  If  y'  >  0  at  the  left*  of  this  point  and  y'  <0  at  the 
right.,  y  is  a  maximum;  if  y^  <  0  at  the  left  atid  y'  ">  0  at  the 
right.,  y  is  a  minimum.  Or.,  if  y"  <iO  at  the  point,  y  is  a 
maximum;  if  y"  >  0,  ?/  is  a  minimum. 

^c)  Iri  an  hiterval  where  ^jy"  >  Q,  the  curve  is  concave 
upward;  where  y"  <  0,  the  curve  is  concave  downwc^rd.. 

(c?)  A  point  at  which^['_==3^is  a^^^  inflection.,  pro- 

vided, y"  changes  sign  as  the  curve  ^(^sses  through  the  point. 

(g)  A  point  at  which  both  y'  =^  and  y"  =  0  is  a  maxi- 
mum or  a  minimum  if  y'  changes  sign  as  the  curve  passes 
through  the  point;  it  is  a  point  of  inflection  with  a  horizontal 
tangent  if  y'  does  not  change  sign. 

Example:  Trace  the  curve 

y  =  x(x  —  1)^. 

This  curve  crosses  the  a;-axis  at  (0,  0),  (1,  0).  For 
large  positive  or  negative  values  of  x,  y  is  large  and 
positive.     The  derivatives  are 

y  z=  (^  _  1)3  j^^^x(x-  1)2=  (x  -  1)2(4  X  -  1), 
y"  =  %x  -  1)(4  a:  -  1)  +  4(a;  -  1)2=  6(:r  -  1)(2  x  -  1). 
The  slope  at  (0,  0)  is  —  1 ;  at  (1,  0)  it  is  0. 

The  critical  values  are  a;  =  1,  \.  When  a:=|,  y"  is 
positive ;  hence  (^,  —  2V6 )  ^^  ^  minimum  point.      When 

2;  =  1,  y"  =  0  '  and  the 
test  by  the  second  de- 
rivative fails.  Since  y' 
does  not  change  sign  as 
-X  ^  passes  through  1,  the 
function  has  neither  a 
Pjq  jj  maximum    nor   a   mini- 

mum at  that  point. 
The  second  derivative  vanishes  at   (J,  —  y^g),   (1,  0)  ; 
these  are  points  of  inflection.     The  slope  at  (^,  —  y^g)  is  ^. 
The  curve  is  shown  in  the  figure. 

V       *  That  is,  immediately  at  the  left. 


GEOMETRIC   APPLICATIONS  37 

EXERCISES 

Trace  the  following  curves.  Where  possible,  find  the  points  of 
intersection  with  the  axes,  determine  the  behavior  of  y  for  large 
values  of  x,  find  the  maxima  and  minima  and  points  of  inflection,  and 
draw  the  tangent  at  each  point  of  inflection. 

1.    ^/  =  a:3  -  6  a;2  +  9  a;  +  3.  2.    ?/  =  4  +  3  x  -  a;^. 

/3.    ?/  =  x3  -  3  x^  +  6  a:  +  10.  4.    ^/  =  (x  -  3)2(x  -  2). 

5.   y  =  (l-  ^2)3.  6.    ?/  =  (4  -  x^y\ 

/7.   y={x-  \y{x  +  2)2.  ^.   y  =  a;3  -  3x2  -  9  x  +  5. 

9.   y  =  x^.  10.   y  =  x^. 

11.   y  =x(x -l)(x-2).  12.?/=       ^^^ 


14.    y  = 


x2  +  4  a2 
1 

(1    +X2')2 


15.  Show  that  the  curve  y  = has  three  points  of  inflection 

x2  +  a'^ 

lying  on  a  straight  line.     Trace  the  curve. 

16.  Show  that,  for  the  curve  y  =  x",  where  n  =  2,  3,  4,  •••,  the 
origin  is  a  minimum  or  a  point  of  inflection  according  as  n  is  even  or 
odd. 

34.  Applications  of  maxima  and  minima.  The  theory 
of  maxima  and  minima  finds  application  in  a  great  variety 
of  problems.  In  the  applications  it  is  rarely  necessary  to 
use  either  of  the  tests  of  ,§  33  to  distinguish  between 
maxima  and  minima ;  the  critical  value  that  gives  the 
desired  result  can  usually  be  selected  by  inspection  of 
the  conditions  of  tlie  problem. 

It  frequently  happens  that  the  function  to  be  teste4  for 
maxima  or  minima  can  be  most  simply  expressed  in 
terms  of  two  variables.  When  this  is  done,  a  relation 
between  the  two  variables  must  be  found  from  the  condi- 
tions of  the  problem-  By  means  of  this  relation  one  of 
the  variables  can  be  eliminated,  after  which  the  maxima 
'and  minima  can  be  found  in  the  usual  way.  However,  it 
is  often  more  convenient  not  to  eliminate,  but  to  proceed 
as  in  example  (5)  below. 


38  CALCULUS 

Examples:  (a)  A  box  is  to  be  made  of  a  piece  of  card- 
board 4  in.  square  by  cutting  equal  squares  out  of  the 
corners  and  turning  up  the  sides.  Find  the  volume  of 
the  largest  box  that  can  be  made  in  this  way. 

Let  X  be  the  length  of  the  side  of  each  of  the  squares 
cut  out.     Then  the  volume  of  the  box  is 

(1)  r=<4V-^^.2. 

The  derivative  is 

^={(4  _  2  a;)2-  4  a;(4  -  2  z) 

=  (4-2:r)(4-6x). 
Setting  ^y 


dx 


we  find 


—    9    ^T-     2 


x^z  ov 


3 


Since  V  vanishes  when  a;  =  0  and  again  when  a:  =  2,  it 
must  reach  a  maximum  at  some  intermediate  point;  it 
therefore  follows  without  the  application  of  further  tests 
that  the  critical  value  x  —  ^  gives  the  required  maximum 
volume : 

F„»x.  =  |(4-|)^  =  Wcu.  in. 

The  graph  of  equation  (1)  is  shown  in  the  figure,  the 

F^-scale  being  one  fourth  as  large  as 
the  a?-scale.  Since  by  the  condi- 
tions of  the  problem  x  is  restricted 
to  values  between  0  and  2,  the 
dotted  portions  of  the  curve  have 

no  meaning"  in  the  present  case. 
Fig.  12  &  r 

(5)  Find  the  dimensions  of  the 

largest  rectangle  that  can  be  inscribed  in  a  given  circle. 

Take  the  coordinate  axes  parallel  to  the  sides  of  the 

rectangle.     The  area  of  the  rectangle  is 

(2)  •  A  =  4.xy. 


GEOMETRIC   APPLICATIONS 


39 


This  can  be  expressed  in  terms  of  x  (or  y)  by  means  of 

the  relation 

(3)  x'^y'^a^ 

which  gives 


y  =  Va^  —  a;^, 


^  =  4  x^ 0^  —  2:^. 

From  this  point  the  method  is  the  same 

as  in  {a).  Fig,  13 

The  problem  can  be  solved  without 
eliminating  x  or  y,  as  follows:    Differentiating   equation 
(2)  with  respect  to  x  and  setting  the  derivative  equal  tu 
0,  we  have,  since  ?/  is  a  function  of  x^ 


^  =  4f^^+.Vo, 


or 


dx 

dy 
dx 


dx 


y. 

X 


Differentiating  equation  (3),  we  get 

ax 


or 


dy  _      x 
dx         y 

Equating  values  of  -^,  we  find 

dx 

_  ^  =  _  ^, 
X         y 

whence 

y  =  x'. 

the  maximum  rectangle  is  a  square. 


EXERCISES 

1.   What  is  the  largest  rectangular  area  that  can  be  inclosed  by 
800  yd.  of  fencing? 


40  CALCULUS 

2.  For  a  rectangle  of  given  area,  what  shape  has  the  minimum 
perimeter  ? 

•f^  3.    Find  the  most  economical  proportions  for  a  cylindrical  tin  cup 
of  given  volume.  Ans,  Radius  =  height. 

4.  A  rectangular  field  is  to  be  fenced  off  along  the  bank  of  a 
straight  river.  If  no  fence  is  needed  along  the  river,  what  is  the 
shape  of  the  rectangle  requiring  the  least  amount  of  fencing? 

5.  The  equal  sides  of  an  isosceles  triangle  are  10  in.  long.  Find 
the  length  of  the  base  if  the  area  is  a  maximum. 

6.  Find  the  rectangle  of  maximum  perimeter  inscribed  in  a  given 
circle. 

^'7.  Find  the  most  economical  proportions  for  a  box  with  an  open 
top  and  a  square  base. 

+  8.  Find  the  most  economical  proportions  for  a  covered  box  whose 
base  is  a  rectangle  with  one  side  twice  the  other. 

A  ns.  Altitude  =  |  x  shorter  side. 

•f-  9.  Find  the  dimensions  of  the  largest  right  circular  cylinder  that 
can  be  inscribed  in  a  given  sphere.  Ans.  Diameter  =  V2  x  height. 

10.  In  Ex.  9,  find  the  form  of  the  cylinder  if  its  convex  surface  is 
a  maximum. 

•^11.  Find  the  dimensions  of  the  largest  rectangle  that  can  be  in- 
scribed in  a  given  right  triangle.  Ans.  x  =  ^  a. 

-fl2.  Find  the  most  economical  proportions  for  a  conical  tent  of 
given  capacity.  Ans.  h  =V2  r. 

J^.  A  man  in  a  rowboat  6  miles  from  shore  desires  to  reach  a  point 
on  the  shore  at  a  distance  of  10  miles  from  his  present  position.  If  he 
can  walk  4  miles  per  hour  and  row  3  miles  per  hour,  where  should  he 
land  in  order  to  reach  his  destination  in  the  shortest  possible  time  ? 

Ans.  1.2  miles  from  his  destination. 

14.  A  rectangular  field  of  given  area  is  to  be  inclosed,  and  divided 
into  two  lots  by  a  parallel  to  one  of  the  sides.  What  must  be  the 
shape  of  the  field  if  the  amount  of  fencing  is  to  be  a  minimum  ? 

15.  A  Norman  window  consists  of  a  rectangle  surmounted  by  a 
semicircle.     What  shape  gives  the  most  light  for  a  given  perime'ter? 

A  ns.  Breadth  =  height. 

•f-16.   Find  the  most  economical  proportions  for  a  quart  can. 

Ans.  Diameter  =  length. 


GEOMETRIC   APPLICATIONS  41 

"/---i7.  The  strength  of  a  rectangular  beam  is  proportional  to  the 
breadth  and  the  square  of  the  depth.  Find  the  shape  of  the  strongest 
beam  that  can  be  cut  from  a  log  of  given  diameter. 

A  ns.  Depth  =  V2  x  breadth. 

18.  Find  the  volume  of  the  largest  box  that  can  be  made  by  cutting 
equal  squares  out  of  the  corners  of  a  piece  of  cardboard  6  x  16  in. 
and  turning  up  the  sides.  Ans.  if^^  cu.  in. 

AS.    A  gutter  is  to  be  made  of  a  strip  of  tin  12  in.     \  / 

wide,  the  cross  section  having  the  form  shown  in  the       V  y 

figure.      What    depth    gives   a   maximum   carrying         \      4      / 
capacity?  '  Fig.  14 

20.  Find  the  most  economical  proportions  for  a  cylindrical  cup  of 
given  capacity,  if  the  bottom  is  to  be  three  times  as  thick  as  the  sides. 

21.  Find  the  most  economical  proportions  for  an  A-tent  of  given 
volume,  whose  sides  slope  at  45°  to  the  horizontal. 

22.  Find  the  dimensions  of  the  largest  right  circular  cylinder  that 
can  be  inscribed  in  a  given  right  circular  cone.      Ans.  Altitude  =  \h. 

23.  Solve  Ex.  22  if  the  convex  surface  of  the  cylinder  is  to  be  a 
maximum. 

24.  Find  the  right  circular  cone  of  maximum  volume  inscribed  in 
a  given  sphere. 

25.  Find  the  cone  of  minimum  volume  circumscribed  about  a  given 
sphere. 

26.  The  base  of  a  box  is  a  rectangle  with  one  side  twice  the  other. 
The  top  and  front  are  to  be  made  of  oak,  the  remainder  of  pine.  If 
oak  is  twice  as  valuable  as  pine,  find  the  most  economical  proportions. 

27.  A  wall  tent  12  x  16  ft.  is  to  contain  a  given  volume.  Find  the 
most  economical  proportions.  Ans.  Height  above  eaves  =  2  a/8  ft. 

28.  An  oil  can  is  made  in  the  shape  of  a  cylinder  surmounted  by  a 
cone.  If  the  radius  of  the  cone  is  three  fourths  of  its  height,  find  the 
most  economical  proportions. 

Ans.  Altitude  of  cylinder  =  altitude  of  cone. 

29.  A  cupboard  5  ft.  high  and  having  shelves  1  ft.  apart  is  to  be 
made  from  a  given  amount  of  material.  If  a  front,  but  no  back, 
is  required,  what  shape  gives  the  greatest  amount  of  shelf  room  ? 

Ans.  Width  =:  twice  depth. 

30.  A  silo  is  made  in  the  form  of  a  cylinder,  with  a  hemispherical 
roof;  there  is  a  floor  of  the  same  thickness  as  the  wall  and  roof. 
Find  the  most  economical  shape.  Ans.   Diameter  =  total  height. 


42 


CALCULUS 


31.  Solve  Ex.  30  if  the  floor  is  twice  as  thick  as  the  wall  and  roof. 

Ans.    Height  of  cylinder  =  diameter. 

32.  The  cost  of  erecting  an  office  building  is  $  100,000  for  the  first 
floor,  $  105,000  for  the  second,  1 110,000  for  the  third,  etc. ;  other  ex- 
penses (lot,  plans,  excavation,  etc.)  are  $700,000.  The  net  annual 
income  is  1 10,000  for  each  story.  How  high  should  the  building  be, 
to  return  the  maximum  rate  of  interest  on  the  investment? 

.   Ans.  17  stories. 

35.  Derived  curves.  The  curves  y  z=zf'(^x)^  y  =f"(x), 
y  =f^'^(jc)^  ...  are  called  th.Q  first.,  second.,  third.,  ...  derived 

curves^  corresponding 
^  to  the  curve  y  =f(x)* 

The  relations  be- 
tween the  given  func- 
tion and  its  first  and 
second  derivatives, 
which  have  been 
formulated  analytic- 
ally in  §  33,  are  well 
brought  out  graphi- 
cally by  drawing  the 
original  curve  and  its 
first  and  second  de- 
rived curves.  This 
is  shown  in  Fig.  15 
for  the  curve 


y  = 


X'^ 


x^ 


The  work  should  be 
arranged  with  the 
several  axes  of  ordi- 
nates  lying  in  the 
same  vertical  line. 
It  often  happens  in  practical  work  that  a  function  is 
defined  in  such  a  way  —  for  instance  by  experimental  data 


Fig.  15 


GEOMETRIC   APPLICATIONS  43 

—  that  no  mathematical  expression  for  it  is  known.  If 
then  we  wish  to  examine  tlie  behavior  of  one  of  the  deriv- 
atives, as  is  frequently  the  case,  we  plot  the  graph  of  the 
original  function  from  the  given  data,  and  construct  the 
required  derived  curve  graphically.  The  process  is  ob- 
vious. We  can  measure  the  slope  at  any  point  of  the 
original  curve  ;  the  number  thus  obtained  is  the  ordinate 
of  the  corresponding  point  on  the  first  derived  curve.  In 
this  way  as  many  points  as  desired  may  be  plotted  on  the 
first  derived  curve  and  a  smooth  curve  drawn  through 
them,  after  which  the  second  derived  curve  may  be  con- 
structed in  a  similar  way;   etc. 

EXERCISES 

1.  What  can  be  said  of  the  first  derived  curve 

(a)  in  an  interval  where  the  original  curve  is  <  „  ,,.       >? 

°  i  falling  J 

(b)  in  an  interval  where  the  original  curve  is  concave  \    ^  ^  ? 

[downward] 

(c)  at  a  point  where  the  original  curve  has  a  <      .    .  >? 

[  minimum  J 

(d)  at  a  point  where  the  original  curve  has  a  point  of  inflection  ? 

2.  What  can  be  said  of  the  second  derived  curve 

(a)  in     an     interval     where     the     original     curve      is     concave 
J  upward       |  r, 

\  downward  j 

(b)  at  a  point  where  the  original  curve  has  a  point  of  inflection? 

(c)  in  an  interval  where  the  first  derived  curve  is    i       ^^®    I  v 

■\  falling  J  * 

(d)  at  a  point  where  the  first  derived  curve  has  a  |  ^^[^^cimum  1 

i  minimum  J  ' 

3.  What  can  be  said  of  the  original  curve  at  a  point  where  the 
second  derived  curve  touches  the  a:-axis  without  crossing  it? 

4.  Plot  the  curve  ij  = x^  +  x'^  and  its  first,  second,  and  third 

derived  curves. 

5.  Plot  the  curve  y  =  sin  x,  and  construct  the  first  derived  curve. 
What  well-known  curve  does  the  latter  resemble  ? 


44 


CALCULUS 


6.    Draw  a  smooth  curve,  on  a  large  scale,  through  the  points 
a-    _4    _2        0    2    4      6     8     10     12     14      16    18    20    22    24 


y        0-1-106    10     3      0-1-1-1       1      9    20    35 

and  construct  the  first  and  second  derived  curves. 

7.  The  national  debt  of  the  United  States  at  the  indicated  dates  is 
given  in  the  accompanying  table,  the  unit  being  $100,000,000.  Con- 
struct the  curve  showing  the  rate  at  which  the  debt  has  increased  or 
decreased. 


Date 
Debt 

Date 
Debt 


1850 
0.6 

75 
20.9 


^55 
0.4 

'80 
19.2 


'60 
0.6 

'85 
13.8 


'61 
0.9 

'90 
8.9 


'62 
5.0 

'95 
9.0 


'63 
11.1 

'99 
11.6 


'64 
17.1 

1900 
11.1 


'68 
24.8 

I  '10 

9.9     10.5 


'65 

26.8 

'05 


'70 
23.3 


CHAPTER   V 


DIFFERENTIATION    OF    TRANSCENDENTAL   FUNCTIONS 


I.    Trigonometric  and  Inverse  Trigonometric 

Functions 

36.  Trigonometric  functions.  The  student  is  already 
familiar  with  the  elementary  properties  of  the  trigono- 
metric functions.  They  are  one-valued  and  continuous  for 
all  values  of  the  argument  a:,  except  that  the  tangent  and 

secant   become  infinite    when    a:=±(2?i  +  l)  — ,  the   co- 

tangent  and  cosecant  become  infinite  when  x  =  ±  wtt,  where 
n  is  a  positive  integer.  The  sine  and  cosine,  and  their  re- 
ciprocals, the  cosecant  and  secant,  are  periodic  with  the 
period  2  tt  ^the  tangent  and  cotangent  are  periodic  with 
the  period  tt. 


y  =.  sin  9CX 

Fig.  16 


The  properties  just  mentioned  are  well  exhibited  by  the 
graphs  of  the  various  functions.     The  graphs  of  the  sine, 


y  =  cos  X 

Fig.  17 


cosine,  and  tangent  are  shown  in  Figs.  16-18  ;  the  student 
should  draw  the  graphs  of  the  other  functions. 

45 


46 


CALCULUS 


function  by  the  general  method 
others  can  be  obtained. 


The  derivative 
is  an  important 
aid  in  the  fur- 
ther study  of 
these  functions. 
Since  all  the  func- 
tions can  be  ex- 
pressed in  terms 
of  the  sincj  it  will 
be  sufficient  to 
find  the  deriva- 
tive of  this  one 
from  this  result  all  the 


37.  Differentiation  of  sin  x.  The  derivative  of  sin  x 
may  be  obtained  directly  from  the  definition  of  the  deriva- 
tive (§  15).     We  have 

y  =  sin  X, 
y  +  Ay  =  sin  {x  -\-  Aa;), 

A?/  =  sin  (x  +  A:r)  —  sin  x^ 

t^y  _  sin  {x  +  A^;)  —  sin  x 

Ax  Ax 

Expanding  sin  (^x  +  Ax^  by  the  addition  formula  of  trigo- 
nometry, we  get 

Ay  _  sin  x  cos  Ax  ■+■  cos  x  sin  Ax  —  sin  x 

Ax 


Ax 


By  trigonometry, 


so  that 


(1) 


cos  Aa:  =  1  —  2  sin^  1  Ax^ 

Ay  _  cos  X  sin  Ax  —  2  sin  x  sin^  ^  Ax 

Ax  Ax 

sin  Ax       .          sin  A-  Ax      .     .  . 
=  cos  X  •  — sin  X  '  — r-2 .  sin  ^  Ax. 


Ax 


iAx 


TRANSCENDENTAL  FUNCTIONS 


47 


It  will  be  shown  in  the  next  article  that 

lim?i!l^=l. 

Assuming  this  result  for  the  moment,  we  see  that 

lim  sin  i  Ax 


T       sin  A:r       -, 
iim  — =  1, 


2 


I  Ax 


=  1. 


Ajr->0       Ax 

Hence,  passing  to  the  limit  in  equation  (1),  we  find 

dy  _  d 

dx      dx 

In  some  applications,  it  is  convenient  to  write  this  formula 
in  the  form 


sm  X  =  cos  X. 


dx 


sin  X  =  sm 


x-\ — 

2 


If  u  is  any  function  of  x^  it  follows  from  formula  (5)  of 
Chapter  III  that 

du 


or 
(7) 


d    '  d     . 

—-  sin  u  =  —~  sin  u  •  -— , 

dx  du  dx 


d    .  du       .    ,      , 

—  sm  z/  =  cos  w  —  =  sin    i/  + 


TT\du 


2Jdx 


dx  dx 

Example:  Differentiate  sin  5  x^. 
By  (7),  with  u  =  5x^ 

— •  sin  5  a:^  =  10  a;  cos  5  x^. 
dx 

38.    Limit  of  sin  a/a  as  a  approaches  0.     In  the  differen- 
tiation of  sin  X  we  had  to  make  use  of  the  fact  that 

lim5iE^=l. 

a-^0      a 

This  result  may  be  obtained  as  follows. 

Let  P,  Q  be  two  points  on  a  circle 
such  that  the  chord  PQ  subtends  an 
angle  2  a  <  tt.  As  a  approaches  0, 
the  ratio  of  the  chord  to  the  arc 
approaches  unity  : 

a->o     SLTcPQ  Fig.  19 


48  CALCULUS 

But 

chord  PQ=2r  sin  oe, 

and,  if  a  is  measured  in  radians, 

arcP§  =  2ra. 
Hence 

T      chord  PO      T      2  r  sin  «      t      sin  a      ^ 
lim — ^  =  iim  =  iim =  1; 

a->.o     arc  PQ         a->o      2  ra  a->o      a 

When  a  is  in  degrees,  the  length  of  the  arc  is 

arc  PQ  =2r  ' a, 

^  180 

and    the    formula   for   —  sin  x  is  much  less    simple   than 

dx 

when  radians  are  used    (see    Ex.   26,  p.   50).     For   this 

reason  angles  in  the  calculus  are  always  measured  in  radians 

unless  the  contrary  is  stated. 

39.  Differentiation  of  cosj:,  tanjr,  etc.  The  derivatives 
of  the  other  trigonometric  functions  can  also  be  obtained 
directly  from  the  definition  of  the  derivative,  but  they  are 
more  easily  found  from  (7). 

To  differentiate  cos  x^  we  write 

cos  X  =  sm  [  a;  +  —  ], 

d  d    '     (     ,  nT\  f     ,   TT 

—  cos  x  =  —  sin  [x  -\-  ~]=  cos  [x  -{ — 
dx  dx       \        2j  V        2 

=  —  sin  X. 

If  u  is  any  function  of  rr,  we  find  by  formula  (5)  of 

Chapter  III, 

^o\  d  .        du  f      .   'TT\dU 

(8)  —  cos  w  =  —  sin  w  —  =  cos    M  +  -    — 

^   ^  dx  dx  \        2jdx 

The  remaining  trigonometric  functions  may  be  differen- 
tiated by  expressing  them  in  terms  of  the  sine  and  cosine. 
The  results  are  as  follows  :  ■'""^^-^ 

(1)  —  tan  X  =  sec^  a;, 

dx 


TRANSCENDENTAL  FUNCTIONS  49 

(2)  —  cot  x  =  —  cosec^  x^ 

dx 

(3)  —  sec  X  =  sec  x  tan  x, 

dx 

(4)  — -  cosec  x  =  —  cosec  x  cot  x. 
dx 

If  u  is  aii}^  function  of  x,  we  find  by  formula  (5)   of 
Chapter  III, 

(9)  —  tan  u  =  sec^  u  — , 

dx  dx 

d       .  n     du 

—  cot  u  =  —  cosec^  u  — , 

dx  dx 

d  .  du 

— -  sec  u  =  sec  u  tan  w— -, 
dx  dx 

d  ,      du 

—-  cosec  u  =  —  cosec  u  cot  u  —  • 

dx  dx 

EXERCISES 

1.  Trace  the  curve  //  =  sin  x,  finding  maxima  and  minima  and 
points  of  inflection,  and  drawing  the  inflectional  tangents. 

2.  Proceed  as  in  Ex.  1  with  the  curves 

(a)  y  =  cos  X ;    j^  y  =  tan  x  ;     (c)  y  =  sec  x. 
Differentiate  the  following  functions. 

3.  (a)   sin  2  x ;  j^  cos  - ;  j^)  tan  (tt  +  a:)  ; 

(d)  X  sec  X  ;  (e)  a;^  cot  a:;  (^)   (3  ^  +  1)  cos  3  6; 

(g)    ^;  (A)   sin^o;;      •  (0  cos3  2^. 

Ans.    (h)    —sin-;   (K)    2  sin  a;  cos  a- ;    (/)    — -6  cos^  2  ^sin2  ^. 
x^         X 


4t.   y  =  X  tan  2  x  +  Vl  +  x'^. 

I 4  X 

5.    ?/  =  vl  +  sin  a,-.       *  6.    ?/  =  -; • 

sm  X 

7.  ?/  =  cot^  4  a:.  vl  ns.    —  12  cot^  4  a:  cosec^  4  x. 

8.  ?/  =  sin3  3  e.  9.    r  ::=  sec  (2  ^  +  1). 

Find  ~  in  the  following  cases. 
dx 

10.    Q.O'&ly  —  .r-  +  4.  ,4rlr  3/ sin x  =  cos  2  x. 

E 


50  CALCULUS 

12    y         sin  2  x  13.   y  ^  Vl  +  tan^x. 

^     *   "^      1  +  cos  2  a; 
14.   3/2  ^  sin  2  X.  ^5-   3/' -2/  =  tan  |- 

16.  If  ^  =  sin  a;,  find  ?/",?/'",•••,  2^('*). 

clx     dP'x 

17.  If  a;  =  cos  a>/,  find  — -,   — — -• 

(1  X 

18.  If  a:  =  ^  sin  ^^  +  5  cos  kt,  show  that  — -r  =  —  k'^x. 

dt'^ 

19.  Obtain  each  of  the  formulas  (l)-(4),  §  39. 

20.  From  the  trigonometric  formula  for  sin  {x  +  a),  deduce  by 
differentiation  the  formula  for  cos  (x  +  a). 

21.  Find  the  tangent  and  normal  to  the  curve  y  =  sin  a:  at  x  =  —  • 

22.  Find   tangents   to   the    curve  y  =  tan  x   parallel   to   the   line 

^  =  2x4-5. 

23.  If /(x)  =  cos2  X,  find/"(x),/'"(x),  •••,/(")(a;). 

d^y 

24.  If  V  =  ^  sin  X,  find  —^  • 

^  dx^ 

d^x 

25.  If  X  =  f  cos  kt,  find  — —-  • 

'  dfi 

26.  Show  that  if  x  is  measured  in  degrees,  the  formula  for  the 
derivative  of  sin  x  becomes 

£sina;  =  j|5Cosx. 

27.  Differentiate  cos  x  directly  from  the  definition  of  the  derivative. 

Sill  X 

28.  Writing  tanx  in  the  form  tan  x  = ,  obtain  the  derivative 

cos  X 

of  tan  X  directly  from  the  definition  of  the  derivative. 

29.  Find  the  maximum  rectangle  inscribed  in  a  circle,  using  trigo- 
nometric functions. 

30.  Find  the  rectangle  of  maximum  perimeter  inscribed  in  a  circle. 

31.  Find  the  right  circular  cylinder  of  maximum  volume  inscribed 
in  a  sphere. 

32.  Find  the  largest  right  circular  cone  that  can  be  inscribed  in  a 
given  sphere.  Ans.    V=^ira^. 

33.  A  steel  girder  30  ft.  long  is  carried  along  a  passage  10  ft.  wide 
and  into  a  corridor  at  right  angles  to  tlie  passage.  The  thickness  of 
the  girder  being  neglected,  how  wide  must  the  corridor  be  in  order 
that  the  girder  may  go  round  the  corner? 


TRANSCENDENTAL  FUNCTIONS 


51 


34.  A  wall  8  ft.  high  is  27  ft.  from  a  house.  Find  the  length  of 
the  shortest  ladder  that  will  reach  the  house  when  one  end  rests  on 
the  ground  outside  the  wall. 

40.  Inverse  trigonometric  functions.  The  symbol  arc- 
sin  x^  or  sin~i  a;,  denotes  the  angle  ivliose  sine  is  x: 

y  =  arcsin  x  ii  x=  sin  y. 

That  is,  the  function  arcsin  x  is  the  inverse  (§  26)  of  the 
function  sin  x.     The  graph  of 

1/  =  arcsin  x 
is  as  shown  in  Fig.  20.     It  is  of  course  the  same  as  that 
of  sin  a;,  with  the  coordinate  axes  interchanged;  i.e.  it  is 
the  reflection  of  the  sine  curve  in  the  line  i/  =  x. 

The  functions  i/  =  arccosa;,  ?/  =  arctan2:,  etc.,  are  de- 
fined in  a  similar  way. 

In  §§  41-42  we  consider  only  the  three  principal  func- 
tions arcsin  x,  arccos  x,  arctan  x.  The  other  three  func- 
tions may  be  treated  similarly. 

41.  Restriction  to  a  single  branch.  The  trigonometric 
functions  are  one-valued  :  to  a  given  value  of  the  argu- 
ment there  corresponds  one  and  but  one  value  of  the  func- 
tion. The  inverse  trigono- 
metric functions,  on  the  other 
hand,  are  infinitely  mani/- val- 
ued :  corresponding  to  a  given 
value  of  the  variable  there 
are  infinitely  many  values  of 
the  function.  Geometrically 
this  means  that  a  line  x  =  Xq, 
if  it  meets  the  curve  at  all, 
meets  it  in  infinitely  many 
points  ;  the  truth  of  this  state- 
ment is  evident  from  a  glance 
at  Figs.  20-22. 

Following  the  rule  of   S  5, 

7    ,,  ,,  •^.  ?/  =  arcsin  X 

we  snail  confine  our  attention  -pio  '>o 


-10 


52 


CALCULUS 


y  =  arctan  x 
Fig.  22 


to  a  single  branch  of  each  of  these  functions ;  the  branch 
chosen  is  the  one  drawn  full  in  each  figure.  Thus  in  our 
future  work  the  function  arcsin  x,  for  example,  is  restricted 
to  the  interval 


IT      .  .  .  TT 

—  <  arcsin  x  <  —  > 

2=  =2 


This  means  that 


arcsin  (—  1)  = 


IT 

2' 


IT 


not  - — ;  etc.     Similarly, 


—  <  arctan  x  <  — 

2=  -2 


therefore 


arctan  (—  1)=  —  — ,  arctan  (—  oo)  =  —  — 


etc. 


TRANSCENDENTAL  FUNCTIONS 


EXERCISES 


\ 


In  the  following,  the  restrictions  laid  down  in  §  41  are  assumed  to 
hold. 

1.  Find  (a)  arcsin  |,  (b)  arcsin  (—  l),  (c)  arctan  (—  V.S),  (d)  arc- 
tan  oo,  (e)  arccos  (—  |),  (/)  arccos  (-1). 

2.  Show  that  arcsin  x  +  arcsin  (—  x)  =  0. 

3.  Show  that  arccos x  +  arccos  (—  x)=  tt. 

4.  Show  that 

(a)  arccos  x  =  -  —  arcsin  x ; 

Ch)      arccot  x  = arctan  x  =  arctan  - ; 

2  X 

/  N  1    TT         .   1 

(c)       arcsec  x  =  arccos-  = arcsm  - ; 

(a)   arccosec  x  =  arcsni  -• 

X 

42.    Differentiation  of  the  inverse  trigonometric  functions. 

To  differentiate  the  function 

?/  =  arcsin  a:, 
let  us  pass  to  the  direct  form 

sin  y  =  X, 
Differentiating  by  the  rule  for  finding  the  derivative  of  an 
implicit  function  (§  25),  we  find 

dii     -, 
cos  y^  =1, 

dx 

dy  ^     1^1  ^1 

dx      cos^      Vl-sin2«/       VH^' 


hence 


or 


d  .  1 

arcsin  x  = 


dx  ^i  _ 


X- 


It  should  be  noticed  that  cos  y  is  put  equal  to  Vl  —  sin^  y 
rather  than  —  Vl  —  sin^  y.  This  is  correct  because,  as 
Fig.  20  shows,  the  slope  of  the  curve  y  —  arcsin  x  is  posi- 
tive at  all  points  of  the  branch  that  we  are  considering. 


54  CALCULUS 

In  a  similar  way  we  find 

d  -1 

■arccos  x  = 


dx  VI  -  a;2' 

d        ,  1 

—  arctan  x  — r,  • 

dx  1  +  x^ 

By  formula  (5)  of  Chapter  III  we  find  that  if  u  is  any 

function  of  a:, 

du 

rM\\  d  dx 

(10)  —  arcsin  u  — 


dx  Vl  -  1/2 

du 

d  dx 

arccos  w=  — 


dx  Vl-i^2 

da 
(11)  ^  arctan  w  =     ^ 


dr  1  +  1/2 

While  in  the  above  discussion  we  confine  our  attention 
to  a  single  branch  of  the  function,  it  appears  from  Figs. 
20-22  that  if  we  know  the  slope  at  every  point  of  one 
branch,  we  can  at  once  find  the  slope  at  every  point  of  any 
other  branch. 


EXERCISES 

Find  the  derivatives  of  the 

following  functions. 

1.   y  =  arcsin  2  x. 

2. 

1 

?/  =  arccos  -  • 

^                          X 

3.    y  =  arctan  (1  +  2  a:). 

4. 

y  =  arcsin  Vx. 

5.   y  =  arccot  (2  x  +  5) 2. 

8. 

1 
V  =  arccosec  -—  • 

-^                      2x 

7.   s  =  t  arcsin  3  t. 

p  =  Vl  —  arcsin  v. 

9.    2/ =  (arcsin  x) 2. 
10.    V-        ^ 

11. 

Ans.    2  arcsin  X 

y  =  arctan  i: . 

arctan  x 


TRANSCENDENTAL  FUNCTIONS  55 

^2.   5  =  Vl  -  2t  arccos  V2T.  13.   y  =  z  arctan  4  r. 

14.  V  =  arcsin  —  •  Ans.    -    ■ — -, 

15.  y  =  t"^  SLTcsin--  16.   y  = 

2  V  arcsin  2  x 

17.  ?/  =  arcsin  x  +  arccos  x.     Explain  the  meaning  of  the  result. 

18.  If  y^  sin  x  +  y  =  arctan  x,  find  y' . 

19.  Find  tangents  to  the  curve  y  =  arctan  x  perpendicular  to  the 
line  4  a:  +  3/  =  0. 

20.  Obtain  — •  arccos  x  from  the  relation 

dx 

arccos  x  = arcsin  x. 

2 

21.  Show  that 

arctan  x  =  arcsin 


and  obtain  —  arctan  x  from  this  fact. 
dx 

22.  If  2/  =  arcsin  x,  find  ^-^,  ^— ^« 

dx^    dx^ 

23.  If  ?/  =  arctan  x,  find  — ^,  — ^ . 

dx'^    dx^ 

24.  Show  that 

d  .  -  1        ' 

arccot  x  = 


Vl  +  ; 


rf  1 

—  arcsec  x — 


dx  x^x'^  —  1 

d         ^  -  1 

—  arccosec  x  = 


dx  xVx"^  -  1 

25.  Trace  the  curve  y  =  arccot  x. 

26.  Trace  the  curve  y  =  arcsec  x. 

27.  Trace  the  curve  y  =  arccosec  x. 

-    II.    Exponential  and  Logarithmic  Functions 

43.  Exponentials  and  logarithms.  The  number  a"(a  >  0) 
is  defined  in  algebra  for  all  rational  values  of  n.  In  the  calcu- 
lus it  becomes  necessary  to  attach  a  meaning  to  the  function 

?/  =  a^  (a>0) 

as  X  varies  continuously. 


56 


CALCULUS 


Let  Xq  be  any  given  irrational  number.  It  can  be  shown 
that  when  x  approaches  Xq  passing  through  rational  values, 
the  function  a-^  approaches  a  definite  limit.  This  limit  is 
denoted  by  a^» : 

lim   a^=^a-">. 


X-^Xo 


The  function  a-^  thus  becomes  defined  for  all  values  of  x. 
This  function  is  one-valued  and  continuous,  and  obeys  the 
ordinary  laws  of  exponents,  viz. : 
(1)  a^  '  a^  =  a 


x+t 


(«')' 


a 


xt 


The  inverse  of  the  exponential  function  is  the  logarithm^ 
defined  by  the  statement  that 

y  =  \og„x  '\{  x=^  dJf  (a>l*). 

This  function  is  one-valued  and^eonttnuous^for  all  positive 
values  of  x.  The  number  a  is  called  the  hase  of  the  sys- 
tem of  logarithms. 

The  graph  of  the  function 

where   g  =  2.718..-  (see  §46),  is  shown  in  Fig.  23;  the 

graph  of  its  inverse 

y  =  log^a? 
is  shown  in  Fig.  24. 


y  =  log  X 


Fig.  24 


*  The  assumption  a  >  1  is  introduced  for  simplicity ;  this  condition  is 
satisfied  in  all  cases  of  practical  importance. 


TRANSCENDENTAL  FUNCTIONS  57 

44.    Properties  of  logarithms.     For  convenient  reference 
we  recall  the  fundamental  properties  of  logarithms  : 
( 1 )  log^a://  =  log^x  +  log„?/, 

(^)  loga  -  =  log«2:  -  log„^, 

(3)  logaX''  =  it  log„2;, 

(4)  log^a^  =  X, 

(5)  '  a}''%''=x, 

(6)  log^a;  =  log^a;  •  log„^, 

To  prove  (1),  let 

then  we  must  show  that 

p  =  m  -{-  n. 
Passing  to  the  direct  form,  we  have 

x^  =  a^,  X  =  a"*,  y  =  a", 
so  that 

Hence,  by  (1)  of  §  43, 

p  z=z  m  +  n. 

Formulas  (2)  and  (3)  may  be  proved  in  a  similar  way. 

Formula  (4)  is  merely  a  restatement  of  the  definition  of 
the  logarithm  ;  formula  (5)  is  the  converse.  To  prove 
(5),  set 

and  take  logarithms  to  the  base  a  on  each  side : 

log^a:  =  log„^, 

whence 

t  =  X. 

To  prove  (6),  let  m  =  log^a:  and  n  =  log^a; ;   then  . 

x  =  a"^  —  J". 
If  we  take  logarithms  to  the  base  a  on  each  side  of'  the 
equation 


58  CALCULUS 

it  appears  that 

As  a  special  case  take  x=  a:  the  formula  gives  (7), 

1 


logft^  = 


loga^ 


EXERCISES 

1.  Find  X,  if  (a)  log^^x  =  2,   (b)  log^^x  =  -  h  (c)  log^x  =  4,  (d) 
logi„a;3=  4,  (e)  loga^;  =  0,   (/)  \ogaX  =  1. 

2.  Simplify  (a)  a^os^  (6)  a'^'^ss^   (c)  a^i^s^  (rf)  aSiogx^  (•g)  a'+iogx^ 
(/)  «*"5^°^*,  the  logarithms  being  taken  to  the  base  a  in  each  case. 

Ans.  (b)  i. 

3.  Prove  formulas  (2)  and  (3)  of  §  44. 

4.  Show  that  negative  numbers  have  no  (real)  logarithms, 

5.  Show  that  numbers  between  0  and  1  have  negative  logarithms ; 
numbers  greater  than  1,  positive  logarithms. 

6.  Show  that 

lim   logoa:  =  —  oo  . 

7.  Find  the  inverse  of  the  function 

a^  —  a~* 

y  = • 

a^  4-  a-^ 

8.  For  what  two  values  of  a:  is  a^    =  (a*)   ? 

45.  The  derivative  of  the  logarithm.  To  obtain  the  de- 
rivative of  the  logarithm  we  proceed  by  the  general  method 
of  §  15  : 

7/  =  log„a:, 

y  +  A?/  =  log„(2;  + Arc), 

X  -\-  Ax 


Ay  =  log„(2;  +  Ax)  -  \og^x  =  log„ 

by  property  (2)  of  §  44.     Hence, 

Ax      Ax         \         X  J 


X 


TRANSCENDENTAL  FUNCTIONS  59 

Let  us  multiply  and  divide  by  x  and  then  make  use  of  (3), 
§44: 

^  =  l.flog/l  +  ^ 
C^x      X     l\x         \         X 


=iiogii+^r. 

X  \  X  J 


Hence, 

(1)  f^  =   lim  ^  =  1  lim  log/l  +  ^f 


(2)  =-  log„ 

X 


Aa^V^ 


lim     1  + 

Ax->0  V  X  J 


It    will    appear   in    the    next    article    that    the    limit 
lim  ( 1  4-  - )    exists  and  is  a  number  lying  between  2  and  3. 

This  number  is  denoted  by  the  letter  e  ;    we  shall   find 
later  (Ex.  4,  p.  230)  that 

e=  lim  Tl  + -Y=  2.71828  •••. 


n->-<x)  \  72/ 

X 

Now,  in  the  limit  occurring  in  (1),  let  us  put  — -  =  n. 

Since  x  is  supposed  to  be  different  from  0,  it  follows  that 
when  Ax  approaches  0,  n  becomes  infinite,  and 

Hence,  assuming  for  the  moment  the  existence  of  e,  we 
have  from  (2) 

(3)  ---log„2:  =  -log„e. 

dx  X 

In  case  the  base  a  ^f  the  system  of  logarithm?  is  the 
number  e,  the  numerical  factor  log^  e  in  formula  (3)  re- 
duces to  unity,  and  the  formula  takes  a  particularly  simple 
form.  For  that  reason  logarithms  to  the  base  e  are  used 
almost  exclusively  in  the  calculus. 


60  CALCULUS 

Logarithms  to  the  base  e  are  called  natural  logarithms^  or 
Napierian  logarithms.  In  our  future  work  the  symbol  log  x^ 
in  which  no  base  is  indicated,  will  be  understood  to  mean 
the  natural  logarithm  of  x.     Thus  we  have  from  (3) 

d  ,  1 

—  log  X  =  -' 

dx  X 

By  formula  (5)  of  Chapter  III,  if  u  is  any  function  of  x, 

du 
(aj  1  ax    1 

J-  lOga  U  =  —  ■  l0g„  e, 

ax  u 

and 

^-«N  d  .  dx 

(12)  Si»g«  =  17- 

3xample :     Find  the  slope  of  the  curve 


y  =  log  Vl  -\-  '6  X 

at  the  point  (a:,  y ). 

Let  us  write  y  in  the  form 

3/  =  i^  log  (1  -f  3  x\ 
Then 

^  ~2  '  l  +  3aj      2  +  62:* 
46.    The  limit  e.     It  will  now  be  shown  that  the  limit 

limfl  +  iy^^ 

exists  and  that  e  is  a  number  between  2  and  3.  For  the 
sake  of  simplicity  we  shall  prove  this  result  only  for  the 
case  when  n  becomes  infinite  passing  througli  positive  in- 
tegral values,  referring  for  the  general  proof  to  more 
advanced  texts. 


TRANSCENDENTAL  FUNCTIONS  61 

When  n  is  a  positive  integer,  we  can  expand  the  quantity 

(  1  H-  -  J   by  the  binomial  theorem  : 
\        nj 

V       nJ 

\nj  2 1        \nj  3 !  \nJ 

n(n  —  1^  ■•  (n  —  'n  —  'i  )  fVy 


n !  \n 


i_i  ri-iYi-? 


=  1  +  1+—— + -, 


+ 


('-3-('-^) 


As  n  increases  the  numb^i*  of  terms  in  the  expansion  in- 
creases, and   every  term   (except  the  first  two)  becomes 

larger.     Hence    the    quantity  (1  +  -)    steadily  increases 
with  n. 

On  the  other  hand,  this  quantity  is  always  less  than  3. 
For,  the  7^  -f- 1  terms  in  the  expansion  (1)  are  each  less 
than  (or,  for  the  first  two  terms,  equal  to)  the  correspond- 
ing terms  of  the  series 


2      92  ■  '   2«-i 

Remembering  that,  by  elementary  algebra,  the  sum  of  the 
geometric  progression  (cf.  §  7) 


1+U4+-+  ' 


9         92    '  9n-l 


is  2 ,  we  find 

9n-l' 


1  +  ^  Y  <  3  -  J-  <  3. 
n/  2^-1 


62  CALCULUS 

We  have  now  shown  thatfl  +  -j  steadily  increases 
with  n^  but  never  becomes  greater  than  3.  It  follows  by 
theorem  IV,  §  8,  that  as  n  increases  the  quantity  (1  +  -) 

approaches  a  limit  e  which  is  not  greater  than  3. 

Since,  in  (1),  the  sum  of  the  first  two  terms  is  2  and  the 
succeeding  terms  are  all  positive,  it  follows  that  e  >  2. 
Hence  e  lies  between  2  and  3. 

As  already  stated,  we  shall  see  later  that 

e  =  2.71828  .... 

47.  Differentiation  of  the  exponential  function.  The  de- 
rivative of  the  exponential  function  «^  may  be  found  as 
follows. 

If 

y  =  a^, 
then 

(1)  log„  y  =  x. 

Differentiating    (1)    by  the    rule    for   implicit   functions 

(§  25),  we  find        ^  ^ 

y  ax 

7    =  z-^ —  =  y  loge  a, 
dx      logo  e 

by  (7),  §  44  ;  hence 

d 
—-  a^  =  a^  logg  a. 

dx 

For  the  case  a  =  e,  this  formula  becomes  simply 

dx 
If  w  is  a  function  of  x^  we  have 

—  a"  =  a"  loge  a  ' 

dx  dx 

This  formula,  too,  becomes  simpler  when  a=  e  : 
(13)  -^6^*  =  ^**^". 

dx  dx 


TRANSCENDENTAL  FUNCTIONS 


63 


EXERCISES 

1.  Show  that  common  logarithms  are  transformed  into  natural 
logarithms  by  the  formula 

log'io  X  =  logio «  •  loge  X 
=  0.4343  logex. 

2.  Show  that 

logea;  =  2.3026  l.ogiox. 

3.  By  means  of  a  table  of  common  logarithms,  show  that 

log2  =  0.693,  log  3  =  1.099,   log  5  =  1.609. 

4.  Using  the  results  of  Ex.  3,  find  log  ^,  log  VS,  log  6,  log  0.1, 

log  ^9. 

Find  the  derivatives  of  the  following  functions. 


^5. 

y  =  log  2  X. 

6. 

y  =  log(\  +2:2). 

8. 
10. 

1       (l-xy 
^-^"^1  +  2.- 

'7. 

y  =  log  y/D  —  X. 

9. 

,.logV^. 

y  =  logio  2  X. 

11. 

y  =  logio  (X2  -  1). 

12. 

y  =  log«  a;2. 

13. 

y  =  log  sin  x. 

14. 

y  =  X  log  X. 

15. 

logx. 

^              X 

16. 

?/  =  (1  -  x^)  log  a;. 

17. 

y  =  log^  X. 

18. 

^  =  log  log  x. 

19. 

y  =  log  log  (1  -  x). 

20. 

y  =  e2x. 

21. 

y  =  ex\ 

22. 

y  =  xe-". 

23. 

y  =  e''  log  X. 

24. 

y  =  10-. 

25. 

y  =  2< 

26. 

y  =  ee'. 

27. 

r  =  eo^. 

28. 

r  =  e^  cos  2  ^. 

29. 

y  =  log  cos  2  X. 

30. 

?/  =  arcsin  log  x. 

31. 

y  =  ecoss*^ 

32. 

y  =  log 

33. 

y  =  sin^  e". 

34. 
36. 

tan  ? 

35. 

y  =  Vl  +  log  X. 

y  3=  arctan  c*. 

37. 

y=(il-e^r. 

1      0 

X- 

38. 

y  =  log  e^*. 

39.  liy  =  e    -    ,  find  y. 

40.  If  /  (6)  =  log  log  sin  2  (9,  find  /'  (0) . 


2 


64  CALCULUS 

(ly 

41.  Given  log  (^x  +  y)=  x^  +  y"^^  find  —-- 

42.  liy  =  log  2-,  find  ^/",  t/'",  •••,  y("). 

^   43.    li  y  =  e%  find  ?/",  y'",  •-■,  ?/<^"^.  .4ns.    ?/("■)  =  a"e°^. 

44.  Find  the  inverse  of  the  function  y  =  e^'°^. 

45.  Find  the  inverse  of  the  function  y  =  log  cos  2  x. 

46.  Find  the  tangent  and  normal  to  the  curve  y  =  log  x  at 
(a)  y  =  0;    (b)  y=-^]    (c)  x  =  e\ 

47.  Show  that  the  curve  y  =.  e"  has  a  constant  subtangent.  Hence 
devise  a  simple  geometric  construction  for  drawing  the  tangent  to 
y  —  e''  dX  any  point. 

48.  Show  how  to  draw  the  tangent  to  the  curve  y  =  log  x. 

49.  Find  the  maximum  and  minimum  points  on  the  curve 
y  =  X  log  X.     Trace  the  curve. 

50.  Trace  the  curve  y  =  e 

51.  li  y  =  xe"",  find  y",  y'",  •••,  ?/''\ 

52.  Trace  the  curve  y  =  xe^. 

53.  Find  the  equation  of  a  tangent  to  the  curve  y  =x  log  x  parallel 
to  the  line  3  x  —  2y  =  5. 

64.  In  passing  from  (1)  to  (2),  §  45,  we  make  use  of  the  principle 
that 

lim  (log  x)  =  log  (lim  x). 

From  which  one  of  our  assumptions  concerning  the  logarithm  does 
this  principle  follow  ? 

48.  Hyperbolic  functions.  A  class  of  exponential  func- 
tions of  frequent  occurrence  in  some  applications  are 
known  as  hyperbolic  functions.  They  are  denoted  by'  the 
symbols  sinh  x  (read  hyperbolic  sine  of  a;),  cosh  rr,  and 
tanh  x^  and  are  defined  as  follows  : 

sinh  X  = , 

1-1 

cosh  x= , 

2 

,      1  g^  —  e~^      sinh  X  ' 

tanh  X  = =  — 

gx  _j_  g  X      cosh  X 


TRANSCENDENTAL  FUNCTIONS  65 

The  reciprocals  of  these  are  cosech  x^  sech  x^  and  coth  x 
respectively.  Tables  of  hyperbolic  functions  have  been 
computed ;  see,  for  example,  Peirce's  Short  Table  of  In- 
tegrals (Ginn  and  Co.). 

The  inverses  of  the  hyperbolic  functions  are  called  anti- 
Tiyperholie  functions : 

y  —  sinli~^a;  \i  x=^  sinh  ?/,  etc. 

The  fundamental  properties  of  the  hyperbolic  functions 
are  easily  obtained  from  the  definitions ;  their  derivation 
is  left  to  the  student  in  the  exercises  below. 

EXERCISES 

1.  Show  that 

cosh^  X  —  sinh-  a;  =  1, 

1  —  tanh^  X  =  sech^  x, 

sinh  2  a;  =  2  sinh  x  cosh  x, 
cosh  2  X  =  cosh-  x  -h  sinh'^x. 

2.  Show  that 


— -  sinh  X  =  cosh  x,    -z-  cosh  x  —  sinh  x 
ax  fix 


3.   Show  that 


sinh-i  X  =  log  (a;  +  Vl  +  x-). 

Fundamental  Differentiation  Formulas 

(1)   1=0. 

^«N      tf   ,         dv       du 
^  ^     dx  dx       dx 

^«/x     d  dv 

du       dv 
y— —  M— 

/A\     d_u_    dx       dx 

^^^     dxv  7^         ' 


66  CALCULUS 

^   ^    dxv~     v^dx' 

^^    dx     du'  dx' 

^  ^     dx  dx' 

du 

(6')  |^V^  =  -^, 

(7)  |sin.=  cos.|.sin(.,-|)|, 

(8)  |cos.  =  -sin.|=cos(.  +  |)|, 

(9)  ^tanw=sec2z/^, 
OAT  dr 

du 
(10)     ^  arcsin  w  =      ^ 


fir  Vl  -  1/2 

(11)    -T-arctanw  =  = -^ 

^     ^     dx  1  +  Z/2 

du 
(12)^log„=f, 

(13)   ^e^^e**^. 


MISCELLANEOUS   EXERCISES 

Find  the  derivatives  of  the  following  functions. 

1.  sin3  - .  2.  log  tan  3  x. 

3.  aretana:3.  4.  (1  -  xy-(2  x  +  Sy. 

5.  e'=°»=^.  6.  Vl  -  c6t  X. 

7.  X  arcsin  --  8.  ^ 


2  VI  -  a,-2 

9.   log^sin^.  10.   arctan  (1  -  ^2). 


TRANSCENDENTAL  FUNCTIONS  67 

11.    qos^2  X. 


13.    arcsin  - 


--     V3-4a: 
10.    • 


17.    log  log  cos  X. 


19.   22^. 


21.    X  log  Vl  —  X. 
23.    sin  a:  cos  2  x. 

-.0  i 

25. 


xVx^  +  1 


27. 


(a:  -  1)^ 


12 

(3x2-4)2 

x^+1 

14. 

tan2(l  -  x). 

X                     X 

16. 

€-  —e   ^ 

2 

18. 

Cos2    (--x\ 

\4         1 

20. 

Vsin  x'^.. 

22. 

(e''  -  1)4. 

24. 

arccos  log  x. 

26. 

2               2     3 

(a3  _  j:^)2. 

28 

sin2  2  e 

(1- cos  2^)2 

30. 

logtan(|  +  |). 

(a:2  +  3  a:  +  3)^ 
29.    log(e2x+  1). 

Find  y'  in  the  following  cases. 

31.    sin  (x  +  y)  =  cos  (x  -  y).  32.    e^^/  =  a;  -f  ?/. 

33.    ^^  +  ^  =  3.  34.    a:-?/ =  tan(a:-?/). 

35.    Find  ?/",  if  ay^  =  x^  36.    Find  y'",  if  a:2  -  ?/2  ^  ^2. 

Find  the  slope  of  each  of  the  following  curves  at  the  point  indicated. 

37.  (a:  -  2/)2  =  3  a:  +  4  ?/  -  14  at  (2,  2). 

38.  y  =  log  X  at  the  point  where  y  =  —  2. 

39.  3^  =  e^  (a)  at  the  point  where  y  =  2  ;  (b)  at  the  point  where 
X  =  log  3. 

40.  arcsin  a:  +  a;^  =  0,  at  the  point  a;  =  —  1. 


CHAPTER    VI 
THE  DIFFERENTIAL 

49.  Order  of  infinitesimals.  We  have  found  tliat  in  the 
problem  of  differentiation  the  increments  Ax  and  A?/  are  in- 
finitesimals, with  Ax  as  the  principal  infinitesimal  (§  10). 

An  idea  of  fundamental  importance  in  the  study  of  in- 
finitesimals is  that  of  order.  Let  an  infinitesimal  v  be  de- 
fined as  a  function  of  a  principal  infinitesimal  u.     If 

where  k=^0^  then  u  and  v  are  said  to  be  infinitesimal  of 
the  same  order ;  if  —"- ^^      -    ».-'■— —^„ .. 

lim  !1  =  0, 
«->o  u 

V  is  said  to  be  of  higher  order  than  u.     More  precisely,  if 
a  number  n  can  be  found  such  that 

lim  iL  =.  y^, 

where  A;  =?^  0,  v  is  said  to  be  infinitesimal  of  the  n-th  order 
with  respect  to'u. 

If  u  and  V  are  of  the  same  order,  we  may  write 

V  =  ku  -\-  eu^ 
where  e  is  an  infinitesimal.  It  is  clear  that  when  u  ap- 
proaches 0,  the  term  eu  approaches  0  more  rapidly  than 
does  ku^  so  that  for  small  values  of  u  the  term  ku  is 
numerically  the  larger.  For  this  reason  the  term  ku  is 
called  the  priyieipal  part  of  v. 

Example :   When  the  side  Z  of  a  square  increases  by  an 
amount  Al,  the  area  increases  by  an  amount 
AA  =  {1  +  Aiy-  P  =2lAl+  'Af. 

68 


IM 

M' 

J 

THE  DIFFERENTIAL  69 

If  A^  approaches  0,  AA  does  also. 
The  two  infinitesimals  are  of  the 
same  order,  since 

lim  ^A  =  lim   (2  I  +  Al)  =  2  I. 

^i-X)  Al       AZ->o 

The  principal  part  of  AA  is  2  lAl. 
The  figure  illustrates  the  fact  that 
AA  consists  of  a  term  of  the  first 

Fig.  2o 

and  a  term  of  the  second  order. 

EXERCISES 

1.  What  is  the  increase  AV  in  the  volume  of  a  cube  of  edge  /  when 
the  side  increases  by  an  amount  Al  ?   Show  that  if  Al  is  infinitesimal, 

•AFis  infinitesimal  of  the  same  order,  and  find  the  principal^part^f 
A  V.     Illustrate  by  a  figure.  Ans.  A  T''  =:  3  r^Al  +  3  ZAI'^  +  Al^- 

2.  Of  the  functions  sin  ^.  sec  0,  tan  ^,  1  -  cos  ^,  which  are  infini- 
tesimal with  respect  to  0  as  the  principal  infinitesimal? 

3.  As  the  radius  of  a  right  circular  cylinder  of  given  altitude 
approaches  0,  the  volume  and  the  total  surface  do  likewise.  Show 
that  the  volume  is  infinitesimal  of  higher  order  than  the  total  surface. 

4.  Given  a  right  circular  cylinder  and  a  right  circular  cone  of  the 
same  base  and  altitude,  show  that 

(a)  if  the  altitude  is  infinitesimal,  the  lateral  surface  of  the  cylin- 
der is  infinitesimal  of  a  higher  order  than  that  of  the  cone; 

(h)  if  the  radius  is  infinitesimal,  the  lateral  surf  aces. of  the  cylinder 
and  cone  are  of  the  same  order  and  (for  small  values  of  the  radius) 
the  former  is  approximately  twice  the  latter. 

5.  Is  the  sum  of  two  infinitesimals  itself  infinitesimal?  Is  the 
product?     Is  the  quotient? 

50.  The  differential.  It  follows  from  the  above  defini- 
tion that  Al/  and  Ax  are  in  general  infinitesimals  of  the 
same  order.  For,  the  limit  of  their  ratio  i^_^J/'^3'nd  this 
in  general  exists  and.Js  dii|erentJrom  0.  Further,  the 
principal  part  of  A^  is  evidently  y'Ax.  It  is  easily  seen 
from  Fig.  26  that  the  principal  part  of  A?/  =  QP'  is  QB, 


70 


CALCULUS 


Fig.  26 


the  segment  of  A?/  cut  off  by  the  tangent  at  P.     For,  the 

slope  at  P  is 

,  _  QR  _  QR 
^       PQ      Ax' 
so  that 

QR  =  ^'Ax. 

The  principal  part  of  Ay  (the 
length  QR  in  Fig.  26)  i%  called 
the  differential'^  of  y  and  is  writ- 
-X    ten  dy : 

dy  =  y'  Ax. 

Hence  the  increment  Ay  consists  in  general  of  the  differ- 
ential dy  plus  an  infinitesimal  of  higher  order.  This  is 
illustrated  by  the  example  of  §  49. 

In  particular,  let  y  =  a: ;   then  y'  =  1,  and 
dy  =  dx  =  Ax ; 
i.e.  the  differential  of  the  independent  variable  is  the  incre- 
ment of  the  variable.     We  may  therefore  write 

dy^y'dx. 
Thus  the  differential  of  any  function  is  equal  to  its  derivative 
multiplied  by  the  differential  of  the  independent  variable. 

The  derivative  of  ^  with  respect  to  x  may  now  be  thought 
of  as  a  quotient  —  the  differential  of  y  divided  by  the 
differential  of  x.     This  is  the  reason  for  using  the  symbol 

-^  to  denote  the  derivative.  The  symbol  -^  may  thus  be 
dx  dx 

considered  as  representing  an  actual  division  —  the  ratio 

dy  -i-  dx.     It  must  be  kept  clearly  in  mind,  however,  that 

the  derivative  is  a  certain  limit,  viz. 

^=   lim  ^. 

dx      A^-^-o  Ax 

If  y  =  f  (x)^  instead  of  writing 

dx 
*  In  case  y'  ^  0.     If  y'  =  0,  then  dii  =  0. 


THE  DIFFERENTIAL  71 

we  may,  and  often  do,  write 

dy=f'(x)dx. 

Thus    the    fundamental    formulas    of    differentiation    are 
often  written  in  this  so-called  differential  notation ;   e.g. 

d(x'^')  =  nx^~'^dx^  d(\og  u)  =  — ,  etc. 

u 

Examples:  (a)   If  ^  =  sin  2  0,  then 

dy=2cos2edd, 

(5)  Find  an   approximate    formula   for   the  area  of   a 
narrow  circular  ring. 

The  area  of  a  circle  of  radius  r  is 

A  =  7rr2. 

If  the  radius  be  increased  by  an  amount  Ar,  the  area  is 
increased  by  an  amount  AA  whose  principal  part  is 

dA  =  2  7rr  dr. 
Hence  the  area  A^  of  a  narrow  circular  ring  is  approxi- 
mately the  product  of  the  circumference  *  by  the  width  w  : 

Af=2  irrw. 

EXERCISES 

Find  the  differential  of  each  of  the  following  functions. 

1.  (a)  x2;  (6)  cos^;  (c)  ^^  _  i ;  (^)  log  a;; 

(e)  arcsm?/;    (/)  tan  2  a ;  {g)   ^^— ^;  (^)  sin2y. 

Ans.    (a)  2xdx\  (h)    -sin  6  dO. 

2.  (a)  (l-3a:2)2;  (6)  log  (1  -  cos2  ^)  ;  (c)  ue^; 

(d)  arctane';  (e)  xVa-i-bx;  (/)    —  • 


Vx 


vT 

Z.   y  =  x(l  —  x2)3.  4.   y  ■. 


4:  X 

5.   V  =  u  sin2  u.  Q.   X  =  y  log  y. 

7.  y  =—•  "  8.   s  =  arcsin  (1  —  f). 

*  Either  the  inner  or  the  outer  circumference. 


72  CALCULUS 

ft  cos  6  ^ -  .    , 

9.   r  =  — -^ —  10.    y=e~'^^u\kx. 

11.    F  =  I  irr^.  12.    X  =  t  sin  aL 

13.    y  =(1  +  OL^)  arctan  «.  14.   //  =  cos^2x. 

15.  Find  the  difference  between  dy  and  A?/,  if  ?/  =  a:^.  Draw  the 
figure. 

16.  Proceed  as  in  Ex.  15  for  the  function  y  =  x'^  —  x'^. 

17.  If  y  =  ^x,  find  A?/  and  dy  and  show  geometrically  w^hy  they 
are  equal. 

18.  If  s  =  16  /2  +  25  ^,  find  the  difference  between  As  and  ds  when 
f  =  12  and  A«  =  .02. 

19.  Draw  figures  to  show  that  dy  may  be  equal  to,  greater  than,  or 
less  than  A?/. 

20.  Show  that  the  error  committed  in  using  the  approximate 
formula  of  example  (6),  §  50,  is  irw^.  When  r  =  10  ft.,  what  is  the 
greatest  allowable  value  of  ic  if  accuracy  to  within  5%  is  required? 

Ans.   About  1  ft. 

21.  If  A  is  the  area  of  a  rectangle  one  of  whose  sides  is  twice  the 
other,  draw  a  figure  showing  the  difference  between  dA  and  A^ 
when  the  length  of  the  side  changes  (cf.  Fig.  25). 

22.  If  V  is  the  volume  of  a  cube,  draw  a  figure  showing  the  differ- 
ence between  dV  and  AF  when  the  length  of  the  edge  of  the  cube 
changes. 

23.  Find  an  approximate  formula  for  the  volume  of  a  thin  cylin- 
drical shell  of  thickness /.  Ans.   2irrht. 

24.  Find  an  approximate  formula  for  the  volume  of  a  thin  spheri- 
cal shell.  What  is  the  greatest  allowable  thickness  for  a  radius  of 
5  ft.  if  accuracy  to  1  %  is  required?  Ans.    About  0.6  in. 

25.  Find  approximately  the  volume  of  wood  required  to  make  a 
covered  cubical  box  of  edge  3  ft.,  using  half -inch  boards. 

Ans.   2\  cu.  ft. 

26.  Work  Ex.  25  if  the  dimensions  of  the  box  are  6,  4,  and  2  ft. 

51.  Parametric  equations ;  implicit  functions.  A  curve 
is  frequently  not  determined  by  an  equation  between  x 
and  y^  but  by  two  equations  giving  x  and  y  in  terms  of  a 
third  variable,  or  parameter.  These  equations  are  called 
parametric  equations  of  the  curve. 


THE  DIFFERENTIAL  73 

For  instance,  the  coordinates  of  a  point  moving  in  a 
plane  are  functions  of  the  time  : 

These  two  equations  may  be  considered  as  parametric 
equations  of  the  path.  Again,  the  equations  of  an  ellipse 
in  terms  of  the  eccentric  angle  (f)  are 

x=  a  cos  (f)^  ^  =  b  sin  cf). 

While  it  may  be  possible  to  eliminate  the  parameter, 
thus  obtaining  the  ordinary  cartesian  equation  of  the 
curve,  it  is  often  more  convenient  not  to  do  so. 

When  dealing  with  parametric  equations,  it  is  conven- 
ient to  use  differentials  in  finding  derivatives,  particu- 
larly the  derivatives  of  higher  order.  The  method  is 
illustrated  by  example  (a)  below. 

Differentials  can  also  be  used  conveniently  in  finding 
derivatives  when  the  relation  between  the  variables  is  an 
implicit  one. 

Examples  :     (a)   Find  — ^  and  — ^  when 

dx  dx^ 

x=Zt^  y  =  t^  —  ■^. 
We  have 

,_dgp  =  S  dt^  dy  =  2  t  dt^  — ^  =  ^^ • 

dx       3 

To  find  — ^,  put  (for  convenience)  -^=y'.     Then 
da^  dx 


hence 


dy'  =  I  dt, 


d^y  _  dy'  __  ^dt  _  ^ 
dx^      dx      Sdt      ^' 


(6)  Find  y'  and  y^'  when  x^  -^  y"^  =  a^. 
Differentiating  both  sides  of  the  equation,  we  get 

2  X  dx  +  2  y  dy  =  0,    y'  =  —  -; 

y 


74  CALCULUS 

J  !  _      y  dx  —  X  dy 

dy  x^ 

—  y  +  x-^      ~  y 

yti  ^§¥  ^ ^  ^ y_ 

dx  y'^  y'^ 

_  —  y*^  —  x^  _  _a^ 

-  ^3  yZ 


EXERCISES 

Find  -^  and  — ^  in  the  following  cases. 
dx  dx^ 

1.  (a)  x=t'^,  y  =  t  -^]  (b)  x=  fi  +  1,   y  =  t^', 

(c)  X  =  cos  2  0,  y  =  sin  2  6 ;     (d)  x  =  a  cos^  6,  y  =  a  sin^  6 ; 
(e)  X  =  e"^',  ?/  =  e«  +  1 ; 
(/)  a:  =  a  (cos  ^  +  0  sin  ^),  y  =  a  (sin  ^  —  ^  cos  ^) . 

Ans.    (a)   ^  =  -±- 

^  ^  dx^         4^3 

2.  (a)  2/2  =  4  ax;  (b)  x^  -  y^  =  1 ;      (c)  a:^  +  y~^  =  a^; 
(^)   xt  +  ?/^  =  a^  ;     (e)  x-if  =  y^\       (/)  x^  +  ?/8  =  3  ^(a.^,. 

.  ,  .   r/2y  4a2 

Ans.    (a)  — ^  = • 

dx'^  7/3 

Find   -^,  using  differentials. 

3.  3x33/2  —  x?/ +  ^2 —  ?/  — 5  =  0.  4.    7/ =  cos  (x  -  2/). 

6.  e^'+y'  =  xy.  6.    ^^^  +  ?/2  =  5; 

■       X  +  y 

7.  log  V^MT'  =  X.  Ans.  ^  =  ^'  +  y'-^ . 

dx  y 

8.  xy  —  x^y'^  -\-  b  y  =  5.  9.    x^  —  3  x^  +  xy^  —2/2  =  0. 


CHAPTER  VII 

CURVATURE 

52.    Differential  of  arc.     Let  s  denote  the  length  of  the 
arc  of  the  plane  curve 

counted     from     some    initial 

point    Pq    up   to    the     point 

P :  (x^   ^),    and    suppose    for 

detiniteness    that  s   increases 

as  X  increases.     The  arc  s  can 

be  regarded  as  a  function  of 

ds 
X.     Its  derivative  —    may  be 

found  as  follows  : 


Fig.  27 


As 
i^x 


^__As_   ^  Va2:'  +  A^ 


As 


pp      i^x       pp'  ^x 


where  As  is  the  length  of  the  arc,  PP  the  length  of  the 
chord,  from  P  :  (x,  y)  to  P' :  (x  +  Aa:,  y  +  A?/).  Since,  as 
pointed  out  in  §  38, 

lim  >« 


,  =  1' 


we  have 

(1) 


Ax-^^  PP 


^  =3  lim  f^  = 


V'  Hr 


If  s  increases  as  x  decreases,  then 


Ax 


\AxJ 


75 


76  CALCULUS 

and 

(2)  ^=-Jl+(&f. 

dx  ^        \dxj 

After  squaring  and  clearing  of  fractions,  equation  (1) 

(or  (2)),  becomes 

i.e.  ds  is  the  hypotenuse  of  the  right  triangle  whose  sides 
are  dx  and  c?«/. 

If  the  tangent  to  the  curve  at  P  makes  an  angle  a  with 
OX^  then 

dx     .  dy 

cos  a  =  — ,  sm  a  =  -^  > 

ds  ds 

53.  Curvature.  We  say  in  ordinary  language  that  a 
curve  whose  direction  changes  rapidly  has  great  curvature^ 
or  is  sharply  curved.  Thus  a  circular  arc  is  said  to  have 
greater  curvature  when  the  radius  is  small  than  when  it  is 
large.  This  somewhat  vague  idea  may  be  made  precise  as 
follows. 

Consider,  first,  two  points  P,  P'  on  a  circle,  and  denote 
the  arc  PP'  by  As,  the  angle  between  the  tangents  at  P, 

P'  by  Aa.     The  quotient  —  is  evidently  the  change  in 

the  direction  of  the  curve,  per  unit  of  arc*  ;  it  is  called  the 

curvature  of  the  circle. 

If  now  the  curve  in  question  is 
not  a  circle,  the  direction  of  the 
curve  no  longer  changes  uniformly, 

and    the    quotient   —    represents 

As 

Fig.  28  merely  the  average  curvature  of  the 

arc  As.     But  if  P'  be  made  to  ap- 
proach P  along  the  curve,  so  that  As  and  Aa  approach  0, 

the  quantity  —  in  general  approaches  a  limit  — ,  which 
*  It  is  easily  seen  that,  in  the  case  of  the  circle,  this  quotient  is  constant. 


CURVATURE  )?7 

is  called  the  curvature  at  the  point  P : 

K=  lim  Aa^rfa^ 
As-^o  As      ds 

The  definition  is  of  course  independent  of  the  particular 
coordinate  system  used ;  the  angle  a  is  the  angle  made  by 
the  tangent  at  P  with  a7i7/  fixed  line  in  the  plane  of  the 
curve.  When  the  equation  of  the  curve  is  given  in  car- 
tesian coordinates,  it  is  convenient  to  take  a  as  the  slope- 
angle  of  the  tangent  —  i.e.  the  angle  between  the  tangent 
and  the  a:-axis.  The  curvature  fc  is  then  easily  expressed 
in  terms  of  the  coordinates.     For, 


tan  a  =  — ^  =  ?/', 
ax 


da 


a  =  arctan  y', 


Also,  by  §  52, 


Hence 

(1) 


ds  =  Vl  -f  y'^  dx. 
da  y" 


ds      (1  +  y'2y 

It  is  customary  to  consider  fc  as  essentially  positive,  so 
that,  strictly  speaking,  we  should  write 


K  = 


da  1-'^ 


ds 


(1  +  y'^y 


where  the  symbol  |  a  \  means  the  absolute   or   numerical 
value  of  a. 

It  should  be  noted  that  when  ^'  =  0,  formula  (1)  re- 
duces to 


fC=l/" 


Thus  the  value  of  the  second  derivative  at  any  point  is 
equal  to  the  curvature  at  that  point  when  the  coordinate 
axes  are  so  chosen  that  the  first  derivative  is  0. 


78 


CALCULUS 


54.    Radius  of  curvature.     The  reciprocal  of  the  curva- 
ture is  called  the  radius  of  curvature^  and  is  denoted  by  p : 

_l  _ds      (1+  y'^y  ^ 
^~K~da~        y^      * 

This   quantity    is   also   to    be    considered   as   essentially 

positive. 

If  a  length  equal  to  the  radius  of  curvature  p  at  the 

point  P  be  laid  off  on  the  normal  from  P  toward  the  con- 
cave side  of  the  curve, 
the  extremity  Q  of  this 
segment  is  called  the 
center  of  curvature.  It 
can  be  shown  that  the 
circle  with  radius  p  and 
center  Q  represents  the 
curve  near  P  more 
closely  than  any  other 
circle.  This  circle  is 
called  the  osculating  cir- 
cle., or  circle  of  curvature. 
In  general,  the  circle 

of  curvature  crosses  the  curve  at  P,  as   is   the    case   in 

Fig.  29. 

EXERCISES 

1.  Show  that  the  curvature  of  a  straight  line  is  everywhere  0. 

2.  Show  that  the  radius  of  curvature  of  a  circle  is  the  radius  of 
the  circle. 

Find  the  radius  of  curvature  of  the  following  curves. 

3.  y  =  x'^  (a)  at  any  point ;   (b)  at  the  vertex. 

4.  y'^  =  4:  ax. 


Fig.  29 


4a2 


5.  The  equilateral  hyperbola  2  xy  =  a^  at  (a,  I  a). 

6.  y  =  x^  +  5 x^+  Qx  at  (0,  0). 


Ans.  fVSa. 
Ans.  22.51. 


7.   The  ellipse  ^4-^=1 
a^     b^ 


(aV  +  Mx^) 


CURVATURE  79 

8.    The  hyperbola  ^  -  ^  =  1. 


1 


9.  The  hypocycloid  a:^  +  ?/^  =  a^.  Ans.  ^(axy)^. 

10.  The  ellipse  x  =  a  cos  cf),  y  =  b  sin  <^. 

11.  The  curve  x  =  t'^,  y  —  \  —  t^. 

12.  The  curve  x  =  ^t\  y  =  iM  -  t^.  Ans.  |(1  +  t'^)'^. 

13.  The  catenary  y  =  ^\e''  -i-  e  V  at  the  point  (0,  a).  Ans.  a. 

14.  Show  that  the  curvature  at  a  point  of  inflection  is  0. 

16.  Find  the  point  of  maximum  curvature  on  the  curve  y  =  e'. 

Ans.  (-0.347,0.707). 

16.  At  what  points  of  the  curve  y  =  x^  \s  the  curvature  greatest? 

17.  Plot  the  parabola  a:^  =  4  y  accurately,  on  a  large  scale,  in  the 
interval  from  a;  =  —  |  to  a;  =  f ,  and  draw  the  osculating  circles  at  the 
points  a:  =  0,  X  =  I,  a;  =  1. 


CHAPTER   VIII 
APPLICATIONS  OF  THE  DERIVATIVE   IN  MECHANICS 

55.    Velocity  and  acceleration  in  rectilinear  motion.     If  a 

point  P  moving  in  a  straight  line  describes  equal  spaces 
in  equal  times,  its  motion  is  said  to  be  uniform.  Its  dis- 
tance X  from  the  starting  point  0  is  evidently  proportional 
to  the  time  : 

X  =  v^t. 

The  constant  factor  v^  is  called  the  velocity  of  the  moving 
point ;   it  is  equal  to  the  space  passed  over  per  unit  time. 

If  the  motion  is  not  uniform,  we  introduce  the  idea  of 
velocity  at  a  jpoint  or  instant.     Suppose  that  a  distance  ^x 

including  the  point  P  is  described 

O  P  in  time  M  :  then  the  quotient  — 

Fig.  30  ^'  .  -;  -  ■ 

IS  the  average  velocity  during  that 
interval  of  time.    If  now  A^  approaches  0  in  such  a  way  that 

P  always  remains  in  Aa:,  the  quotient  — -  approaches  a  limit 

which  is  called  the  velocity  at  the  point  P.  This  limit  is 
of  course  the  derivative  of  x  with  respect  to  t  : 

y^  lim  ^^dx 

Thus  the  velocity  is  the  time-rate  of  change  of  space 
described  (cf.  §  6). 

The    rate    of    change    of    the    velocity    is    called    the 
acceleration : 


j= 

~  dt 

80 

d'^x 
dt^ 

APPLICATIONS  OF  THE  DERIVATIVE  81 

If  the  acceleration  y  is  constant,  the  motion  is  said  to  be 
nn'^ormly  accelerated.  An  important  case  of  uniformly 
ac^celerated  motion  is  that  of  a  body  falling  toward  the 
earth  from  a  point  near  the  earth's  surface,  all  resistances 
being  neglected.  The  attraction  of  the  earth  gives  the 
body  an  acceleration  ^,  called  the  acceleration  of  gravity^ 
equal  to  32  ft.  per  second  per  second'approximately. 

Tn  any  problem,  it  is  instructive  to  draw  the  graphs  of 
^  V;j^  and^^^as  functions  of ^^^^  In  doing  this,  it  should  be 
remembered  that  the  graph  of  v  is  the  first  derived  curve 
(§35),  the  graph  ofy  is  the  second  derived  curve,  cor- 
responding to  the  graph  of  x. 


EXERCISES  * 

1.  A  stone  is  thrown  upward  with  a  velocity  of  64  ft.  per  second. 
The  distance  from  the  starting  point  at  the  time  t  (in  seconds)  is 

y  =  16  r^  -  64  U 
the   positive    sense   being   downward.     Find    the    velocity    and    the 
acceleration.     How  high  will  the  stone  rise  and  for  how  long  a  time  ? 
Where  is  the  stone  and  what  is  its  velocity  after  5  seconds  of  motion? 
What  distance  is  covered  in  the  sixth  second  V 

2.  In  Ex.  1,  draw  the  graphs  of  _?/,  i\  and  /. 

3.  A  particle  slides  down  an  inclined  plane.      The  distance  from 
the  starting  point  at  any  time  t  is 

a—  4  f2  _  20  t. 

Discuss  the  motion. 

4.  A  point  moves  according  to  the  law  a;  =  5  cos  2  t.     Discuss  the 
motion.     Draw  the  graphs  of  x,  \\  and  j. 

5.  A  point  moves  according  to  the  law  ^  =  32  (1  —  e-*).     Discuss 
the  motion. 

6.  A  point  moves  according  to  the  law  x  —  log  (1  +  2  t).     Discuss 
the  motion.     Draw  the  graphs  of  x,  i\  and/. 

7.  A  point  moves  according  to  the  law  x  —  e~'  sin  2  t.     Discuss  the 
motion. 

*  The  types  of  motion  considered  here  will  be  discussed  more  in  detail 
in  Chapter  XXVII. 

G 


82 


CALCULUS 


8.    The  positions  of  a  point  at  the  ends  of  successive  seconds  are 
observed  as  follows : 


t\0       1 


3     4     5     6     7 


X\0      -1       -A 


32 
'3 


Fig.  31 


-1     0     I     4     7 

Draw  the  graphs  of  v  and  J,  and  find  an  approximate  expression  for 
V  andy  in  terms  of  t.  ■  ,  ' 

9fi^  Vectors.  A  right  line  segment  of  definite  length, 
direction,  and  sense  is  called  a  vector.  Vectors  are  of  great 
importance  in  physics  because  they  can  be  used  to  repre- 
sent velocities,  accelerations,  forces,  and  other  fundamental 
quantities. 

The  resultant  of  two  vectors  AB,  ^C^  is  the  diagonal 
AD  of  the  parallelogram  having  AB,  AC  SiS  adjacent  sides. 

Two  forces  acting  on  the  same  par- 
ticle are  equivalent  to  a  single  force, 
their  resultant ;  similarly  for  other 
vectors.  This  is  the  parallelogram 
law.  The  operation  of  finding  the 
resultant  by  the  parallelogram  law  is  called  geometric 
addition. 

The  original  vectors  AB,  AC  are  called  components  of 
AD.  It  is  evident  that  any  vector  may  be  resolved  into 
cetpappnents  in  an  infinite  number  of  ways. 

sV'i.,  Velocity  in  curvilinear  motion.  If  a  moving  point 
describes  a  plane  curve,  its  coordinates  are  functions  of 

the  time  : 

x=<j>{t),  7/  =  ylr(t). 

The  distance  s  passed  over  along  the 
curve  is  also  a  function  of  the  time. 
The  velocity  at  any  point  P  is 
defined  as  the  vector,  laid  off  on 
the  tangent  to  the  path  from  P,  of 
magnitude 


y^   lim  As^Js 

At^O  ^f       fit 


Fig.  32 


APPLICATIONS  OF  THE  DERIVATIVE  §3 

^'\ 

The  components  of  the  velocity  parallel  to  the  coordi- 
nate axes  are 

Vj.  =  v  cos  a,  Vy=  V  sin  a, 

where  a  is  the  angle  between  OJTand  the  tangent  at  P. 

By  §  52, 

ds      dx      dx        •  ds      dv      dy 

V  cos  «  =  —  •  —  =  — ,  V  sin  a  =  —  •  -^  =  -^ , 

dt      ds      dt  dt      ds       dt 

so  that  , 

dx  dy 

dt     ^     dt 

Thus  the  rectangular  components  of  the  velocity  of  P 
are  the  velocities  of  the  projections  P^  and  Py  of  P  on 
the  coordinate  axes. 

By  §  ^Q^  the  total  velocity  v  is 

inclined  to  the  ic-axis  at  an  angle 

V   * 
a  =  arctan  -^  • 

The  equations 

a:  =  </)(0,  y  =  ir(t) 

may  be  regarded  as  parametric  equations  (§  51)  of  the 
path  in  terms  of  the  parameter  t.  The  cartesian  equation 
may  be  obtained  by  eliminating  the  parameter. 

58,  Rotation.  If  a  point  moves  in 
a  circle  at  a  uniform  rate,  so  that 
equal  angles  are  swept  out  in  equal 
times,  the  angle  6  swept  out  by  the 
radius  vector  in  the  time  t  is 
6  =  od^t. 

The  constant  w^  is  called  the  angular  Fig.  33 

velocity. 

If  the  motion  is  not  uniform,  we  are  led  as  in  §  55  to 
define  the  angular  velocity  at  a  particular  instant  as 

CD  = . 

dt 


'^^  CALCULUS 

Since  ds  =  r  dO,  where  r  is  the  radius  of    the  circle,  it 
follows  that  the  linear  velocity  v  =  —   and  the   angular 

velocity  co  are  connected  by  the  relation 

V  =  cor. 

The  rate  of  change  of  the  angular  velocity  is  called  the 
angular  acceleration^  and  is  denoted  by  a  : 

dt      df ' 


EXERCISES 

1.  A  man  can  row  a  boat  5  mi.  per  hour.  Tf  he  pulls  at  right 
angles  to  the  course  of  a  river  2  mi.  wide  having  a  current  of  3  mi. 
per  hour,  where  and  when  will  he  reach  the  opposite  shore  ? 

2.  In  Ex.  1,  if  the  man  wishes  to  land  directly  opposite  his  start- 
ing point,  in  what  direction  must  he  row  and  how  long  will  it  take 
him  to  cross?  ^ns.  30  min. 

3.  A  steamship  is  moving  at  the  rate  of  12  mi.  per  hour.  A  man 
walks  across  the  deck  at  right  angles  to  the  ship's  course,  at  the  rate 
of  5  mi.  per  hour.  If  the  deck  is  40  ft.  wide,  how  far  is  he  finally 
from  his  starting  point  ? 

4.  If  a  point  moves  so  that 

X  =  a  cos  t,  y  =  a  sin  t, 

find   the   total  velocity   in  magnitude  and  direction    at  the  time  t. 
What  is  the  path  described  ? 

5.  Find  the  path  and  discuss  the  motion  of  a  point  whose  co- 
ordinates are 

x  =  3t,  y  =  t  —  7. 

6.  The  equations  of  the  path  of  a  moving  body  in  terms  of  the 

time  are 

x  =  20t,  y  =  16  fi. 

Find  the  position  of  the  body,  its  distance  from  the  starting  point, 
and  the  magnitude  and  direction  of  the  velocity  when  t  =  2. 

7.  A  flywheel  2  ft.  in  diameter  makes  100  revolutions  per  minute. 
Find  its  angular  velocity  in  radians  per  second,  and  the  linear  velocity 
of  a  point  on  the  rim.  What  constant  angular  retardation  (negative 
acceleration)  would  bring  it  to  rest  in  10  seconds? 


APPLICATIONS  OF  THE  DERIVATIVE 


8.  A  point  moves  in  a  circle  in  such  a  way  that 

^  =  4  f2  _  3  ^ 
Find  (u  and  a,  and  draw  the  graphs  of  0,  w,  and  a  as  functions  of  t. 

9.  Find  the  angular  velocity  in  Ex.  4. 

10.    In  Ex.  8,  find  Vj.  and  Vy  when  t  =  1  if  the  radius  of  the  circle  is 
10  ft.  Ans.   v^  =  -  42.1,  Vy  =  27.0  ft.  per  sec. 

59.  Acceleration  in  curvilinear  motion.  Suppose  the 
velocity  of  the  moving  point  P  at  the  time  t  is  v,  at  the 
time  t  -\-  At  is  v'  =  V  +  Av^  where  Av  is  the  vector  which, 
geometrically  added 
to  V,  produces  1)'.  If 
V  and  v'  be  laid  off 
from  a  common  ori- 
gin 0,  the  third  side 
W  of  the  triangle 
(Fig.  35)  is  evi- 
dently Av.  Now  as 
A^  approaches  0,  Av 
does   likewise ;    but  Fig.  34 

in  general  the  ratio 

Av 

approaches  a  definite  limit,  and  the  direction  of  Av  ap- 
proaches a  definite  limiting 
direction. 

The  vector  of  length 

Av 


J 


lim 

A<->0  A^ 


laid  off  in  the  limiting  direc- 
tion of  Av,  is  called  the  ac- 
celeration of  P  at  the  time  t. 
It  is  the  so-called  geometric 
derivative^  or  vector  deriva- 
tive, of  v  with  respect  to  t. 
To  find  an  expression  for  j  in  terms  of  the  coordinates 
of  P,  we  may  resolve  j  into  components  jj.  and  jy  parallel 


,< 


CALCULUS 


to  the  coordinate  axes.     Denoting  by  <^'  the  angle  between 

Av  and  the  2;-axis,  let  us  project  the  triangle  OFF'  on  OX  : 

Av  cos  </)'  =  vj  —  Vj.  =  AVj.. 

Dividing  by  Af,  we  get 

Av         ,,     Av^ 
—  cos  (p'  =  — -. 

At        ^       At 

whence,  in  the  limit  when  A^  approaches  0, 

where  <^,  the  limiting  value  of  </>',  is  the  angle  between  j 
and  the  a:-axis.     Similarly 

.   .      .       dv^      d?x 
1  sin  <h  =  — ^  =  -—. 

-^        ^       dt       dt^ 
Thus 


7«=: 


dV:, 


dt 


d^ 
dt^' 


*'''      dt      df ' 


The  total  acceleration  j  is 


inclined  to  the  a^-axis  at  an  angle 


<f)  =  arctan*^. 

Jx 


It  is  often  more  conven- 
ient to  resolve  j  into 
components  j^  and  j^  along 
the  tangent  and  the  nor- 
mal to  the  curve  at  JP. 
These  components  can  be 
found  directly  from  the 
definition  of  J  ;  for  vari- 
ety, liowever,  we  will  find 

them    by    projecting   the   components  j^   and  jy   on  the 

tangent  and  normal. 


Fig.  37 


APPLICATIONS  OF  THE  DERIVATIVE  J^^ 

If  the  tangent  at  P  makes  an  angle  a  with  the  2;-axis, 
then 

jt  =  jx  cos  a  +  jy  sin  a 

_  dv^     dx      dVy     dy 

dt      ds       dt      ds 

dv^     dx      dVy     dy 

dt      dt       dt      dt 


ds 
dt 

dv_  dVy 

^  dt  ^    '  dt 


V 

But,  differentiating  the  equation 
with  respect  to  ^,  we  find 


» 

dv 

dt~ 

o     dv.  ,   o     dVy         dv^  ,       dv^ 

2V«>,2  +  t,„2                               V 

Hence, 

.      dv 

Again. 

) 

Jn  =  Jv  COS  a  -  j^  sin  a 
_  dVy    dx     dv^    dy 
dt      ds       dt      ds 

By  §  54, 

the  radius  of  curvature  of  the  path  is 

ds 
P=da' 

so  that 

p\dt      da       dt      da  J 

1  f    dVy         dv^ 
p\  ^ da         ^  da 
Now,  differentiating  both  sides  of  the  equation  (§  57) 

a  =  arctan  -^ 


88  CALCULUS 

with  respect  to  a,  we  find 


dv„  dv 

1 

1  = 


V 

1  da  da 


so  that 


Hence, 


V  2 

dv^  dv^ 

V —  V 

da  da 

dv^  dv^        n 

V —  V  — -  =  v^. 

da  aa 


i;2 


Jn  — 

P 

Thus  the  acceleration  j  is  equal  to  —  07dy  in  the  case  of 

rectilinear   motion;    in    curvilinear    motion  —    represents 

■         dt 
the  tangential  component  oi  the  acceleration. 

EXERCISES 

1.  In  Ex.  4,  p.  84,  find  y^,  y^,  y^,  y„.  Find  y,  (a)  as  the  resultant 
of  /j.  andy^^,  (&)  as  the  resultant  oi  jt  andy,i. 

2.  In  Ex.  6,  p.  84,  find  the  total  acceleration  in  magnitude  and 
direction  when  t  =  2. 

3.  Show  that  in  uniform  circular  motion  the  acceleration  is 
directed  toward  the  center  and -is  proportional  to  the  radius  of  the 
circle. 

4.  In  Ex.  8,  p.  85,  find  ji  and  y„,  if  the  radius  of  the  circle  is 
10  ft. 

5.  A  point  describes  the  parabola  ^/^  =  4  x  with  a  constant  velocity 
of  6  ft.  per  second.     Find  v^,  Vy,  j\,  andy^  at  the  point  (1,  2). 

/  60.  Time-rates.  The  question  of  determining  time- 
rates  arises  in  a  variety  of  problems  beside  those  that  have 
been  considered. 

If  in  any  problem  the  quantity  whose  rate  gf  cjiange  is 
to  be  found  can  be  expressed  directry  as  a  function  of 
the  time,  the  result  can  of  course  be  obtained  at  once  by 


APPLICATIONS  OF  THE   DERIVATIVE 


89 


differentiating  with  respect  to  the  time.  Frequently, 
however,  the  problem  is  solved  by  expressing  the  quantity 
in  question  Jii  tgiuns  of  anotlier  quantity  whose  rate  of 
change  is  known,  and  then  differentiating  the  equation 
connecting  them.     The  methodTis  best  explained  by  an 

JExample :  Water   is   flowing   into    a   conical   reservoir 
20  ft.  deep  and  10  ft.  across  the  top,  at  the  rate  of  15  cu. 
ft.  per  minute.     Find  how  fast  the  surface  is  rising  when 
the  water  is  8  ft.  deep. 
'     The  volume  of  water  is 


r=  i  irr^h. 

By  similar  triangles. 

r       5 
A~20'   *■" 

v- 

Hence 

7rA3 

48' 

dV='^}Uh, 
16 

dV      irh^dh 

dt        16  dt 

But  we  have  given  that 

^^=15, 
dt 

so  that 

ttA^  dh  _^r     dh 
16   dt~      '     dt 

240 

7rA2 

When  A  =  8, 

dh       15       , 

-i  f\    £j^ 

=  —  =  1.19  ft.  per  minute. 

dt  4:77  ^ 


EXERCISE 

1.  Water  is  flowing  into  a  cylindrical  tank  of  radius  5  ft.  at  the^ 
rate  of  20  gallons  per  second.     Find  how  fast  the  surface  is  rising. 

2.  In  the  example  of  §  60,  find  how  fast  the  water  is  flowing  in 
if,  when  the  water  is  5  ft.  deep,  the  surface  is  rising  2  ft.  per  minute. 


90  CALCULUS 

3.  Water  is  flowing  into  an  inverted  conical  tank  32  ft.  deep 
and  12  ft.  across  at  the  bottom,  at  the  rate  of  4  cu.  ft.  per  second. 
How  fast  is  the  surface  rising  ? 

4.  Two  trains  start  from  the  same  point  at  the  same  time,  one 
going  due  east  at  the  rate  of  40  mi.  per  hour,  the  other  north  60  mi. 
per  hour.     At  what  rate  do  they  separate ?      A7is.   72.1  mi.  per  hour. 

5.  Two  railroad  tracks  intersect  at  right  angles.  At  noon  there 
is  a  train  on  each  track  approaching  the  crossing  at  40  mi.  per  hour, 
one  being  100  mi.,  the  other  200  mi.  distant.  Find  (a)  how  fast 
they  are  approaching  each  other,  (b)  when  they  will  be  the  nearest 
together,  and  (c)  what  will  be  their  minimum  distance  apart. 

Ans.    (b)  3  :  4.5  p.m.  ;  (c)  70.7  mi. 

6.  A  ladder  20  ft.  long  leans  against  a  vertical  wall.  If  the 
lower  end  is  being  moved  away  from  the  wall  at  the  rate  of  2  ft.  per 
second,  how  fast  is  the  top  descending  when  the  lower  end  is  12  ft. 
from  the  wall  ? 

7.  A  man  6  ft.  tall  walks  away  from  a  lamp-post  10  ft.  high  at  the 
rate  of  4  mi.  per  hour,  (a)  How  fast  is  the  further  end  of  his  shadow 
moving?     (b)  How  fast  is  the  shadow  lengthening? 

8.  A  man  on  a  wharf  20  ft.  above  the  water  pulls  in  a  rope,  to 
which  a  boat  is  attached,  at  the  rate  of  4  ft.  per  second.  At  what 
rate  is  the  boat  approaching  the  shore  when  there  is  25  ft.  of  rope 
out? 

9.  A  kite  is  120  ft.  high,  with  130  ft.  of  cord  out.  If  the  kite 
moves  horizontally  4  mi.  per  hour  directly  away  from  the  boy  flying 
it,  how  fast  is  the  cord  being  paid  out  ? 

10.  A  stone  dropped  into  a  pond  sends  out  a  series  of  concentric 
ripples.  If  the  radius  of  the  outer  ripple  increases  steadily  at  the 
rate  of  6  ft.  per  second,  how  fast  is  the  disturbed  area  increasing  at 
the  end  of  2  seconds?  Ans.  452  sq.  ft.  per  sec. 

11.  The  path  traced  by  a  moving  point  is  the  parabola  y  =  x^ 
+  2x4-3.     If  y^  =  3  f  t.  per  second,  find  Vj^  and  the  total  velocity  v. 

Ans.   Uj,  =  6  a:  +  6. 

12.  A  point  moves  on  the  hyperbola  x^  —  y^  =  144  with  v^  =  15  ft. 
per  second.     Find  v  at  the  point  (13,  5). 

-f»  13.  As  a  man  walks  across  a  bridge  at  the  rate  of  5  ft.  per  second, 
a  boat  passes  directly  beneath  him  at  10  ft.  per  second.  If  the  bridge 
is  30  ft.  above  the  water,  how  fast  are  the  man  and  the  boat  separat- 
ing 3  seconds  later?  Ans.   8i  ft.  per  sec 


APPLICATIONS  OF  THE  DERIVATIVE  91 

14.  A  light  is  placed  on  the  ground  30  ft.  from  a  building.  A 
man  6  ft.  tall  walks  from  the  light  toward  the  building,  at  the  rate  of 
5  ft.  per  second.  Find  the  rate  at  which  his  shadow  on  the  wall  is 
shortening  when  he  is  15  ft.  from  the  building.         Ans.   4  ft.  per  sec. 

-15.   Solve  Ex.  14  if  the  light  is  8  ft.  above  the  ground. 

16.  An  elevated  train  on  a  track  30  ft.  above  the  ground  crosses  a 
street  at  the  rate  of  20  ft.  per  second,  at  the  instant  that  an  auto- 
mobile, approaching  at  the  rate  of  30  ft.  per  second,  is  40  ft.  up  the 
street.  Find  how  fast  the  train  and  the  automobile  are  separating 
2  seconds  later. 

17.  In  Ex.  16,  find  when  the  train  and  the  automobile  are  nearest 
together.  Ans.   ^f  sec. 

18.  A  light  stands  60  ft.  from  a  building.  A  man  walks  along  a 
path  20  ft.  from  the  building,  at  the  rate  of  5  ft.  per  second.  How 
fast  does  his  shadow  move  on  the  building? 

19.  An  arc  light  hangs  at  a  height  of  30  ft.  above  the  center  of  a 
street  60  ft.  wide.  A  man  6  ft.  tall  walks  along  the  sidewalk  at  the 
rate  of  4  ft.  per  second.  How  fast  is  his  shadow  lengthening  when 
he  is  40  ft.  up  the  street?  Ajis.   0.8  ft.  per  sec. 

20.  In  Ex.  19,  how  fast  is  the  tip  of  the  shadow  moving? 

21.  A  light  stands  30  ft.  from  a  house,  and  20  ft.  from  the  path 
leading  from  the  house  to  the  street.  A  man  walks  along  the  path  at 
5  ft.  per  second.  How  fast  does  his  shadow  move  on  the  waU  when 
he  is  20  ft.  from  the  house  ? 


Xi 


CHAPTER   IX 

CURVE   TRACING   IN  CARTESIAN   COORDINATES 

I.   Algebraic  Curves 

61.  Introduction.  In  Chapter  IV  we  learned  how  to 
trace  simple  curves  whose  equations  are  given  in  the  ex- 
plicit form 

and  for  which  y,  y\  and  ^"  are  one-valued  and  continuous. 
In  the  present  chapter  we  shall  attempt  a  more  general 
treatment  of  the  subject  of  curve  tracing. 

In  §§  62-67  we  confine  our  attention  to  algebraic  curves 
—  i.e.  curves  for  which  the  ordinate  y  is  an  algebraic  func- 
tion of  X. 

62.  Singular  points.  If  y  is  defined  implicitly  as  a 
function  of  x  by  the  equation 

F(x,  y)  =  0, 

the  derivative  in  general  takes  the  form  of  a  fraction 
whose  numerator  and  denominator  are  functions  of  both 
X  and  y  :   say 

,  ^  A(x,  y) 
^       Bix,  y) 

If  A(x.,  y')  and  B(x^  y^  both  vanish  at  the  point  P  :  (x^  y) 
on  the  curve,  tlie  slope  at  that  point  assumes  the  inde- 
terminate form  -.     A  point  at  which  the  derivative  takes 

the  form  -  is  called  a  singular  point. 

To  find  the  singular  points  of  a  curve  we  must  therefore    j 

92 


CURVE  TRACING  IN  CARTESIAN  COORDINATES  ^^S 

find  the  values  of  x  and  y  that  satisf}^  the  three  equations 

Fix,  y)  =  0, 
A(x,  y)=0, 
B(x,  y)=0. 

As  we  have  but  two  unknowns  x  and  y  to  satisfy  three 
equations,  it  appears  that  a  curve  will  have  singular 
points  only  under  certain  conditions. 

It  will  be  sufficient  to  consider  an  algebraic  curve 
having  a  singular  point  at  the  origin.  If  a  singularity 
occurs  at  any  other  point  (A,  ^),  the  origin  may  be  trans- 
ferred to  that  point  by  the  substitutions 

X  =  x^  -\-  h^ 
y  =  y^  +  k. 

63.  Determination  of  tangents  by  inspection.  Let  the 
equation  of  the  curve  be  written  in  the  form 

F(ix,  y)=  Uq  +  V  +  ^1^+  V^  +  V^  +  (^2^+  •••  +  ^n^"=  ^■ 
Differentiating,  we  find 

{bQ-\-2cQX-{-  c^y  +  •..)c?2:  +  (^i  +  c^x -\-  2c\^y  -i, )dy  =  0, 

dy^_  6q+  2cqX  +  ^1^  +  '" 

dx         5j  +  c-^x  +  2  c^y  +  '" 
The  origin  is  on  the  curve  only  if  a^  =  0.     In  that  case 
the  equation  of    the  tangent  at  (0,  0)  is    found    by  the 
usual  method  (§  27)  to  be 

h^x  +  h^y  =  0, 

provided  b^  and  b^  are  not  both  0  ;  i.e.  the  equation  of  the 
tangent  at  tlie  origin  may  be  found  by  simply  equating  to 
0  the  group  of  terms  of  the  first  degree. 

In  case  a^,  6^  and  b-^  are  all  0,  the  origin  is  on  the  curve 
and  the  derivative  is  indeterminate  at  that  point;  hence  the 
origin  is  a  singular  point.  In  this  case  the  equation  of 
the  curve  evidently  contains  no  terms  of  lower  degree 
than  the  second. 


94  CALCULUS 


For  convenience  let  us  put^ 


=  ^2(^  -  m^x)iy  -m^x). 
Then 

The  abscissas  of  the  points  of  intersection  of  the  line 

y  =  mx 
with  this  curve  are  given  by  the  equation 

Two  roots  of  this  equation  are  0 :  every  line  y  =  mx  inter- 
sects the  curve  in  two  coincident  points  at  the  origin. 
But  the  above  equation  in  x  also  shows  that  if  we  let  m 
approach  either  m^  or  ^2,  the  coefficient  of  oc^  approaches  0 ; 
i.e.  a  third  point  of  intersection  of  the  curve  with  the  line 
y  =  mx  approaches  the  origin,  and  the  lines 

y  =  m^x^    y  =  m^ 

are  both  tangent  to  the  curve  at  the  singular  point.     These 
lines  may  of  course  be  real  and  distinct,  real  and  coincident, 
or  imaginary. 
Since 

H^y  -  ^1^)  (y  -  ^2^) = v^  +  H^y  +  ^2^^ 

we  see  that  the  equations  of  the  two  tangents  are  obtained 
by  equating  the  second  degree  terms  to  0,  and  factoring  the 
left  member  of  the  resulting  equation. 

The  argument  we  have  used  can  be  extended  to  show 
that  if  F{x^  ?/)  has  no  terms  of  degree  lower  than  the  A;th, 
any  line  through  the  origin  meets  the  curve  there  in  k 
points,  and  the  k  tangents  to  the  curve  at  the  origin  are 
obtained  by  equating  the  group  of  terms  of  lowest  degree 
to  0. 

*  The  argument  is  readily  modified  to  take  care  of  the  case  c^  —  0. 


CURVE  TRACING  IN  CARTESIAN  COORDINATES  ^5 

64.  Kinds  of  singular  points.  A  point  at  which  there 
are  two  tangents  (whether  distinct,  coincident,  or  imagi- 
nary) is  called  a  double  point;  one  at  which  there  are 
three  tangents  is  a  triple  point ;  etc.  It  follows  from  §  63 
that  the  origin  is  a  double  point  if  the  equation  F(x^  y)  =  0 
has  terms  of  the  second,  but  none  of  lower,  degree;  a  triple 
point  if  the  equation  has  terms  of  the  third  but  none  of 
lower  degree ;   etc. 

If  the  tangents  at  a  double  point 
are  real  and  different,  the  point  is 
called  a  node :  two  branches  of  the 
curve  cross  each  other,  as  in  Fisf.  39. 

Ti-    ^1         .  .  .  ,  Fig.  39 

It    the    tangents    are    imaginary,  the 
point  is  called  an  isolated  or  conjugate  point :  there  are  no 
other  points  of  the  curve  in  its  vicinity.     Such  a  point  is 
F  in  Fig.  40. 

If  the  tangents  are  real   and   coincident, 
there  are  several  possibilities.     The  simplest 
singularity  in  this  case  is  the  cusp  of  the 
first  kind :  two  branches  of  the  curve  touch 
each  other,  coming  up  on  opposite  sides  of 
Fig.  40     '     ^^6  cuspidal  tangent,  as  in  Fig.  41.     At  a 
cusp  of  the  second  kind  the  two  branches  lie 
on  the  same  side  of  the   tangent,  as   in    Fig.   42.     Fre- 
quently the  point  is  a  double  cusp^  or  point  of  osculation, 


Fig.  41  Fig.  42  Fig.  43 


the  commonest  form  of  which  is  shown  in  Fig.  43.     And 
in  some  cases  the  point  may  be  an  isolated  point. 


96  CALCULUS 

Example  :  Examine  the  curve  y^  =  7?  —  x^  for  singular 
points. 

Since  there  are  no  terms  of  lower  than  the  second  degree, 
the  curve  has  a  singular  point  at  the  origin.  The  tangents 
at  that  point  are  given  by  the  equation 

'if'  =  —  7?'  \ 
i.e.  they  are  the  lines 

y  =  ix, 

y  —  —  ix. 
These  lines  are  imaginary,  and  the  origin  is  an  isolated 
point. 

There  are  no  other  singular  points.     For, 

,       Zx^  —  ^x 

and  the  coordinates  of  the  singular  point  must  satisfy  the 
three  equations 

Sx^-2x  =  0, 
2y  =  0, 
y^  z=  a^  —  x^. 

The  first  two  equations  are  satisfied  by  the  coordinates 
(0,  0),  (|,  0),  but  the  second  pair  do  not  satisfy  the  last 
equation.     (See  also  Ex.  20  below.) 

EXERCISES 

Show  that  the  origin  is  a  singular  point  for  each  of  the  following 
curves,  write  the  equations  of  the  tangents  there,  and  determine  the 
nature  of  the  singularity. 

1.  The  folium  x^  +  ^^  _  3  f^j.^^  ^^^j    Xode. 

2.  x'^y^  =  a^(x'^  +11'^)- 

3.  y 


x+  y 


4.    The  eissoid  ?/  — ^^j^.  Cusp  of  the  first  kind. 

2  a  —  x 

6.    y^  =2  ax'^  —  x^. 

6.  y^{x^  +  y-)  =  a^x'^.  Ans.    Double  cus]). 

7.  y\x-  —  a^)  =  x^.  Ans.  Isolated  point. 


CURVE  TRACING  IN  CARTESIAN  COORDINATES 

8.  {y  —  x^Y  =  x^.  Ans.  Cusp  of  the  second  kind. 

9.  y'^  =  x^  +  x^. 

10.  x^  —  xy'^  =  y^.  Ans.  Triple  point. 

11.  The  leniniseate  Qx~  -\-y^)'^z=  a^(x^  —  y^). 

12.  Show  that  the  conchoid  x^y'^  =  (a  +  yy~(b'^  —  y"^)  has  a  node  at 
(0,  -  a)  iib>a. 

13.  Show  that  the  curve  a(y  —  xy-=(x  —  ay  has  a  cusp  at  (a,  a). 
Find  the  singular  points  of  the  following  curves. 

14.  if  =x(x  -  ly. 

15.  2/2  ^  a:(2  x  +1)2. 

16.  y^  =  x(x^  —  !)•  ^^5-  None. 

17.  ay^  =  x\x  -  ay.  Ans.  Cusps  at  (0,  0),  (a,  0). 

18.  y^  +  y^=  (x^  -  1)2. 

19.  xy^  +  x^y  =  a^. 

20.  Prove  that  a  cubic  curve  cannot  have  more  than  one  singular 
point. 

21.  Prove  that  the  graph  of  a  one-valued  algebraic  function 

cannot  have  any  singular  points. 

65,  Asymptotes.  As  the  point  of  contact  of  a  tangent 
to  a  curve  recedes  indefinitelj;.^from  the  origin,  the  tan- 
gent may  or  may  not  approach  a  limiting  position.  If  it 
does,  the  line  approached  is  called  an  asymptote. 

For  example,  the  hyperbola 

has  the  lines 

a      0 

as  asymptotes.  On  the  other  hand,  the  parabola  has  no 
asymptotes,  since  as  the  point  of  tangency  recedes  the 
tangent  does  not  approach  any  limiting  position. 

Although  general  methods  for  finding  asymptotes  exist, 
they  are   frequently   difficult   to   apply.      The    following 
.tests  are  sufficient  in  ordinary  cases. 


98  CALCULUS 

(a)  To  find  the  conditions  that  must  be  satisfied  in  or- 
der that  the  line 

y  —  mx  4-  k 

shall  be  an  asymptote  to  the  algebraic  curve 

(1)  F{x,  y)  =  a^x^  -}-  a^x'^-'^y  +  ...  +«„?/»  +  h^^-'^ 

+  V"~^«/+  •••  =  ^' 

let  us  substitute  mx  +  k  for  y  in  the  equation  of  the  curve. 
This  gives  an  equation  of  the  nth.  degree  in  x  whose  roots 
are  the  abscissas  of  the  n  (real  or  imaginary)  points  of  in- 
tersection of  the  line  with  the  curve. 

It  is  shown  in  algebra  that  one  root  of  the  equation 
A^x""  +  A^x""-^  -f  ^2^"~^  +  . . .  H-  J.^  =  0 
becomes  infinite  if  Aq  approaches  0  ;  two  roots  become  in- 
finite if  both  Aq  and  A^  approach  0  ;   etc.      Hence  if  we 
equate  to  0  the  coefficients  of  x^  and  2;"~\  we  shall  in  gen- 
eral determine  values  of  m  and  k  such  that  the  line 

y  =  mx  +  k 

will  intersect  the  curve  in  two  infinitely  distant  points. 
Such  a  line  is  in  general  an  asymptote.  Of  course  if  the 
coefficients  of  x"^  and  x^~^  cannot  both  vanish,  there  are  no 
asymptotes  (except  such  as  may  be  given  by  (6)  below). 

Example :  («)  Find  the  asymptotes  of  the  hyperbola 
x^-y'^-2x-2y-[-l==0. 

Substituting  y  =  mx  -f-A:,  we  get 

x^  -  (mx  +  k)^-  2x  -  2(mx  +  A:) -|-  1  =  0, 
or 

(1  -  m2)2:2H-(-  2mk-2m-2^x+  ...  =  0. 

Equating  to  0  the  coefficients  of  the  two  highest  powers  of 
a;,  we  find 

1  -  w2  =  0, 

Whence 

w  =  1,  ^  =  —  2, 
m  =  —  1,  A;  =  0, 


CURVE  TRACL^TG  IN  CARTESIAN  COORDLVATES     99 


Fig.  44 


and  the  asymptotes  are 
y  =  x-2, 
y  =  -x. 

The  curve  is  shown  in  Fig.  44. 

(5)  Asymptotes  parallel  to  the 
?/-axis  are  not  given  by  test  (a), 
since  their  equations  cannot  be  writ- 
ten in  the  slope  form. 

Let  us  arrange  the  equation  of  the  curve  in  descending 
powers  of  y: 

F{x,  y)  =  a^-  +  (a^x  +  ;5i)?/-i 

+  (a^x^  +  ^^x  +  72)r"^  +  ...  =0. 

If  the  term  in  y^  is  present,  every  value  of  x  gives  n  finite 
(real  or  imaginary)  values  of  ?/,  and  no  line  x=  k  can  in- 
tersect the  curve  in  infinitely  distant  points.  But  if 
a^  =  0,  then  every  line  x  =  k  intersects  the  curve  in  one 
infinitely  distant  point.  If  now  the  coefficient  of  y'^~^ 
involves  x  (i.e.  if  a^  ^  0),  that  coefficient  equated  to  0 
gives  us  the  equation  of  a  line  parallel  to  the  ?/-axis 
which  intersects  the  curve  in  tivo  infinitely  distant  points, 
and  is  an  asymptote. 

This  result  can  be  extended  to  the  case  when  the  co- 
efficient of  the  highest  power  of  ?/  is  a  polynomial  of  higher 
degree  in  x.  By  equating  this  polynomial  to  0  we  find 
an  asymptote  parallel  to  01^  corresponding  to  each  real 
linear  factor  of  the  polynomial. 

Similarly,  asymptotes  parallel  to  OX  may  be  found  by 
equating  to  0  the  coefficient  of  the  highest  power  of  x. 
Such  asymptotes  are  detected  in  general  by  test  (a),  but 
it  may  be  easier  to  find  them  by  the  present  method. 

Example :  (5)  Test  the  curve  (x^  —  V)y'^  =  2^  for 
•asymptotes. 

Equating  to  0  the  coefficient  of  the  highest  power  of  ?/, 


100  CALCULUS 

we  find  the  lines 

x^  —  1  =  0,  i.e.  X  —  ±\ 

as  asymptotes  parallel  to  OY.  The  coefficient  of  the 
highest  power  of  x  cannot  be  equated  to  0,  so  there  are  no 
asymptotes  parallel  to  OX. 

To  test  for  asymptotes  oblique  to  the  axes,  put  y  =  mx  +  k  : 

{x^  —  V){if)^7p'  -f  2  mhx  +  A;2)  =  7?. 

Equating  to  0  the  coefficients  of  the  two  highest  powers  of 
x^  we  find 

m^  =  0, 
2  mk  —  1. 

These  equations  are  incompatible ;  hence  there  are  no 
oblique  asymptotes. 

EXERCISES 

Test  the  following  curves  for  asymptotes . 

1.    xy  -\-  X  =  b. 

'  2.    x^  -{-  y^  =  \.  Ans.    X  ■\-  y  =  Q. 

x^ 

3.    y^  =  ax^  +  x^.  4.    The  cissoid  y'^  = 

2  rt  —  X 

5.  x^  +  7  xy  +  12  y^  +  X  -\-  4  y  -  16  =  0. 

Ajis.    a;  +  3  ?/  +  1  =  0,  x  -\-  "1  y  =  0. 

6.  x"^  —  xy  +  y^  -\-  5  X  =  0.  Ans.   None. 

7.  x'^y'^  =  n^(x'^  +  ?/-).  8.    The  folium  x^  +  y^  —  3  axy. 
9.    x'^y'^  +  lx'^+  ?/-  +  X  =  0.      10.    x~y  +  xy'^  =  a^. 

11.  x3  -  4  xy-^  -  3  x^  +  12  xy  -  12  ;/2  +  8  x  +  2  ?/  +  4  =  0. 

A71S.    a;  +  3  =  0,  X  -  2  ?/  =  0,   r  +  2  ?/  =  6. 

12.  ay^  —  ay'^  —  x^  -\-  ax^  +  a^.  .4ns.    x  =  a,  x  ±  y  -\-  a  =  0. 

13.  a?/2  =  x^  +  xy^. 

14.  Prove  that  a  parabola  has  no  asymptotes,  but  that  every  line 
parallel  to  the  axis  meets  the  curve  in  one  infinitely  distant  point. 

15.  Prove  that  every  line  parallel  to  an  asymptote  meets  the  curve 
in  one  infinitely  distant  point. 

16.  In  example   (h),  §  65,  prove   that  every  line   parallel   to   the* 
ar-axis  meets  the  curve  in  one  infinitely  distant  point. 


CURVE  TRACING  IN  CARTESai^,N  ,COQ;RD.INAT£S  IQl 

17.  Show  that  a  curve  of  the  n-th  degree  cannot  have  more  than  n 
asymptotes. 

18.    Show  that  the  curve 

y  =  P(^), 

where  P(x)  is  any  polynomial  in  x,  has  no  asymptotes. 

66.  Exceptional  cases.  Exceptions  may  arise  to  the 
theory  of  §  65.  For  instance,  it  may  happen  that  the  co- 
efficient of  x'^~^  vanishes  identically  for  some  value  of  m  for 
which  the  term  in  x""  disappears,  so  that  all  lines  having 
this  value  of  m  as  their  slope  meet  the  curve  in  two  points 
at  infinity.  In  this  case  there  are  in  general  two  or  more 
parallel  asymptotes  having  the  given  slope,  and  the  values 
of  k  are  determined  by  equating  to  0  the  coefficient  of  the 
highest  power  of  x  that  does  not  disappear  identically. 

The  exceptional  cases  are  rare  and  unimportant  in 
elementary  work,  and  a  fuller  discussion  of  them  is  un- 
necessary. 

EXERCISES 

1.  Show  that  the  curve 

(a;  +  y)'^(x^  +  xy  -\-  y'^)  =  a^y-  +  cfi{x  —  y) 
has  the  pair  of  parallel  asymptotes  x  +  y  =  ±  a. 

2.  Show  that  every  line  parallel  to  the  x-axis  meets  the  curve 

y^  =  x^  -{-  X 
in  two  infinitely  distant  points,  but  that  the  curve  has  no  asymptotes. 

67.  General  directions  for  tracing  algebraic  curves.     The 

following  questions  should  be  answered  as  fully  as  possible 
before  trying  to  trace  an  algebraic  curve. 

(1)  Is  the  curve  symmetric  with  respect  to  the  coordinate 
axes?  (It  is  symmetric  with  respect  to  01^ if  the  equation 
is  unchanged  when  x  is  changed  to  —  rr  ;   etc.) 

(2)  Where  does  it  intersect  the  axes  ? 

(3)  Has  it  any  asymptotes?  If  so,  locate  each  of  the 
points  where  the  curve  intersects  its  asymptotes. 

(4)  Is  it  possible  to  determine  certain  regions  of  the  plane 
within  which  the  curve  must  lie? 


102  CALCULUS 

(5)  Has  the  curve  any  singular  points  ?  If  so,  determine 
the  tangents  at  each  point,  and  the  nature  of  the  singu- 
larity ;   draw  the  tangents  if  they  are  reaL 

(6)  Has  it  any  maximum  and  minimum  points  ? 

The  above  is  only  a  general  outline  of  the  process  to  be 
followed  ;  other  steps  will  often  suggest  themselves.  In 
many  cases  the  points  of  inflection  should  be  found  and 
the  inflectional  tangents  drawn,  but  this  is  not  worth  while 
if  the  second  derivative  is  complicated.  Translation  or 
rotation  of  axes  is  occasionally  useful.  The  elementary 
method  of  tracing  the  curve  by  plotting  points  is  too  la- 
borious to  be  used  extensively,  but  it  is  frequently  advis- 
able to  plot  a  few  points,  merely  as  a  check  on  the  analysis. 

Examples  :  (a)  Trace  the  curve  ?/^  =  3  ax^  —  x?. 

(1)  The  curve  is  not  symmetric  with  respect  to  either 
axis. 

(2)  When  a:  =  0,  ^  =  0  ;   when  ?/  =  0,  a;  =  0  or  3  a. 

(3)  By  §  ^b,  the  line 

y  ^  a  —  x 

is  an  asymptote.  Substituting  a  —  x  for  y  in  the  equa- 
tion of  the  curve,  we  find 

a^  —  3  c^x  +  3  a:i^  —  rr^  _  3  ^^2_  ^3^ 
The  highest  powers  of  x  drop  out,  as  they  should,  and 

we  find  that  the  curve  crosses  its  asymptote  at  a;  =  - . 

o 

(4)  Writing  the  equation  in  the  form 

?/3  =  x^{Z  a  —  a:), 

we  see  that  1/  >  0  if  a:  <  3  «,  ^  <  0  if  a:  >  3  a.  Hence  the 
curve  is  above  the  a:-axis  at  the  left  of  a;  =  3  a,  below  at 
the  right  of  that  point. 

(5)  Since  there  are  no  terms  of  lower  than  the  second 
degree,  the  origin  is  a  singular  point.  The  tangents  are 
given  by  the  equation 

3  aa;2  =  0  : 


CURVE  TRACING  IN  CARTESIAN  COORDINATES  103 


they  coincide  in  the  ?/-axis.  Since,  by  (4),  the  curve 
near  the  origin  cannot  go  below  OX^  the  point  is  a  cusp. 
For  small  values  of  x  on  either  side  of  0,  y  is  real,  hence 
there  is  a  branch  on  each  side  of  OY^  and  the  origin  is  a 
cusp  of  the  first  kind. 

Since  the  curve  is  a  cubic,  there  can  be  no  other  singu- 
lar points  (Ex.  20,  p.  97). 

(6)   The  derivative  is 

dy  _Q  ax  —  Sx^ 
dx  3  ?/2 

The  numerator  vanishes  when  x  =  0  or  2  a.  Rejecting 
the  value  a;  =  0,  which  gives  the  singular  point,  we 
have  a;  =  2  a  as  the  only  critical  value.  It  will  appear 
presently  that  the  point  (2  a,  V4  a)  is  a  maximum  point. 
To  trace  the  curve,  let  us  begin  at  the  extreme  left. 
In  that  region,  the  curve  must  be  just  below  its  asymp- 
tote, since  it  has  to  pass  through  the  origin  and  can  cross 


a 


the   asymptote    only  at   rr=-.       It    comes   down  to   the 


a    2  a\       T,    . 

-,  —  ).      It   IS    now 


origin    tangent    to   the    j/-axis,    turns    back  on  the  other 

side,  and  crosses  the  asymptote  at 
clear  that  the  critical  value 
x=2  a  corresponds  to  a 
maximum  point.  Return- 
ing from  the  maximum,  the 
curve  crosses  OX  at  (3  a,  0) 
and  again  approaches  the 
asymptote. 

The    curve    is    shown    in 
the  figure. 

(5)    Trace    the    curve 
ap'y'^  =  x^^cp'  —  dtp'). 

(1)   The  curve  is  symmet- 
ric   with   respect    to    both  axes. 


Fig.  45 


104  CALCULUS 

(2)  When  a;  =  0,  ?/  =  0  ;   when  7/  =  0^  x  =  0,'±  a. 

(3)  There  are  no  asymptotes. 

(4)  When  x  is  numerically  greater  than  a,  1/^  is  negative 
and  ?/  is  imaginary.  Hence,  the  curve  lies  entirely  be- 
tween the  lines  x  =  ±  a. 

(5)  The  origin  is  a  singular  point.  The  tangents  are 
real  and  different, 

hence  the  point  is  a  node.     There  are  no  other  singular 
points. 

(6)  The  first  derivative  is 

dy  _  2  a^x  —  4a^  _  x(^a^  —  2  x^^ 
dx  2  ^  y 

This  vanishes  when 

«2  _  2  a;2  =  0,  a:  =  ±  -^ . 

V2 

Corresponding  to  each  of  these  values  the  curve  has,  on 

account  of  its  symmetry,  a  maximum  y  =■-  and  a  mini- 

a 
mum  %i  = 

^  2 

The  student  may  draw  the  curve. 

EXERCISES 

Trace  the  following  curves. 

\-x 
1  +  a; 

a'-x  A     ,„  _  a.-^  +  «' 


1.  ^  =  rT-r2-  2.  y  =  x{x  -  1)2. 


{x  —  a)'-^ 


5.   ?/2  = 


4-  .r  ^    yi  ^x^Cl  +  a,-), 


^ 


1  -a: 


7.    ?/  =  2  x3  -  9  a-2  +  12  a;  -  3.  8.    ?/  =  a:^  -  3  a:^  +  0  a:. 

9.    ?/2(a;2  +  «2)  ^  ^2^2.  10.    ?/(a:2  _  ^2)  ^  ^  _^  X. 

11.    a;3  +  y3  3,  3  aa-y.  12.    3/2  ^  3,(3,2  _  i). 


13.   3/  =  -/^-  14.    3/^  = 


a: 


a;2  —  a^ 


16. 

,_x^(x'^-n^) 
■^           x^  +  a2 

18. 

(a:^  —  a'^)y'^  =  ax^. 

20. 

xy"^  +  a;2?/  =  a^. 

22. 

xhf  =  (a;  +  a)2(4  a2  _ 

-x^). 

24. 

(^  —  x^)^  =  x^. 

26. 

ay^  =(a;2  -  a2)2. 

28. 

V-       ^^        . 

CURVE  TRACING  IN  CARTESIAN  COORDINATES     105 

15.   3/2(a;2  +  r/2)  =  arx'^. 
17.    (^  -  a;)2  =  a:3. 

19.  y=i^--y. 

21.   a;2/  =  o:^{x'^  +  y^). 

23.   (a:2  -  ?/2)  (a:  -  3  ?/)  =  x. 

25.    y2  ^  1  -  a:  . 

^        1  +  a;2 

27.    :.3^  =  «2(^  +  «)2.  ,^        ^^^^^^ 

II.    Transcendental  Curves 

68,  Tracing  of  transcendental  curves.  In  tracing  tran- 
scendental curves,  we  follow  much  the  same  procedure  as 
m  §  67,  except  in  the  matter  of  as3"mptotes  and  singular 
points.  While  the  definitions  of  §§  62-65  hold  for  all 
curves,  the  tests  there  given  apply  only  to  algebraic 
curves. 

We  shall  in  this  article  confine  our  attention  to  tran- 
scendental curves  having  no  singular  points.  To  find 
asj^mptotes,  the  following  rule  may  be  used  : 

In  general^  if  y  becomes  infinite  as  x  approaches  a  definite 
limit  a,  the  line  x=  a  is  an  asymptote  ;  if  x  becomes  infinite 
as  y  approaches  5,  the  line  y  —  b  is  an  asymptote. 

In  rare  instances,  the  derivative  may  behave  in  such  a 
way  that  although  the  conditions  of  the  rule  are  satisfied, 
the  tangent  does  not  approach  any  limiting  position,  and 
hence  there  is  no  asymptote ;  but  the  rule  holds  in  all 
cases  that  are  apt  to  arise  in  practice. 

Example  :  Trace  the  curve  y  =  xe^. 

(1)  There  is  no  symmetry. 

(2)  The  curve  crosses  the  axes  at  (0,  0). 

(3)  As  X  becomes  large  and  negative,  y  approaches  0  *  ; 

*  This  statement  is  easily  made  plausible ;  a  strict  proof  will  be  given 
later  (§  140). 


CALCULUS 


hence  the  negative  x-axis  is  an  asymptote.     When  x  is 
large  and  positive,  y  is  large  and  positive. 

(4)  Since  e^"  is  always  positive,  y  has  always  the  same 
sign  as  x :  the  curve  lies  in  the  first  and  third  quadrants. 

(5)  Since   y'  =  xe""  +  e%   the    only 

critical  point  is  [  —  1, ].    The  slope 

at  (0,  0)  is  1.    ^  ^^ 

(6)  y''=  xe^  -\-  2e^.        There    is    a 

point  of  inflection  at  f  —  2, j ;  the 

slope  of  the  inflectional  tangent  is 
-e-2  =  -0.14. 
The  curve  is  shown  in  Fig.  46. 


Fig.  46 


EXERCISES 

Trace  the  following 

curves. 

1.    y  =  e-^\ 

2.   y  =  xe-^"". 

3. 

y  =  tan  x. 

^.   y  =  sec  X. 

-            logx 

X 

6. 

y  =  e'\ 

7.   y  =  sin^  x. 
10.  2/2  =  ^°§'^. 

X 

8.  y'^  =  sin  x. 

11.    2/2  =  log  X. 

9. 
12. 

X 

y  = 

logx 

y  =  -' 

X 

13.  y=,-^' 

logx 
16.    Show  that 

14.   y  =  g-^  sin  x. 
Urn    2  +  sin.^^O; 

16. 

y  =  X  log  X. 

then  show  that  the  curve 


2  +  sin  x^ 


furnishes  an  exception  to  the  rule  of  §  68. 

^'  69.    Curve   tracing   by   composition   of    ordinates.     The 

curve 

can  be  traced  very  readil)^  U  f  (x)  has  the  form 


CURVE  TRACING  IN  CARTESIAN  COORDINATES  3w^ 
where 

are  curves  whose  form  is  easily  obtained.  We  have  only 
to  add  the  ordinates  of  the  two  latter  curves  to  obtain  the 
required  curve. 

Less  frequently  curves  may  be  conveniently  traced  by 
multiplying  or  dividing  ordinates  in  a  similar  way. 

W^  EXERCISES 

Trace  the  following  curves. 

1.  y  =  X  +  \ogx.  2.   y  =  X  -\ 

X 

Z.   y  —  e^  —  X.  ^.   y  ■=  sin  x  +  cos  x. 

6.   y  =  X  -\-  sin  X.  Q.    y  =  sinh  x  = • 


„            sin  X                                              o            cos  X 
7.   y  = 8.   y  = 

X  X 

9.  The  catenary  is  the  curve  in  which  a  homogeneous  cord  or  chain 
hangs  when  suspended  from  two  of  its  points  under  its  own  weight. 
The  equation  is 

X      a  I    ^  ^\ 

y  =  acosh-=-f  ea_,.g-aj. 

Ttace  the  curve. 

70.    Graphic  solution   of   equations.     The    roots    of   the 

equation 

are  the  abscissas  of  the  points  where  the  curve  y=f{x) 
crosses  the  2:-axis.  Hence  if  we  trace  the  curve  y=zf{x) 
and  measure  its  intercepts  on  OX^  we  have  a  graphic 
solution  of  the  equation /(a;)  =  0.  It  is  usually  best  to 
get  the  general  form  of  the  curve  by  the  methods  of  the 
preceding  articles,  and  then  plot  it  carefully,  on  a  large 
scale,  in  the  neighborhood  of  each  of  its  a;-intersections. 

The  roots  of  the  equation 
(1)  f  {X-)  =  <f,ix-} 

are  the  abscissas  of  the  points  of  intersection  of  the  curves 

y  =  </>(^)- 


108 i:/  CALCULUS 


In  case  these  two  curves  are  easily  traced,  we  thus  ob- 
tain with  little  labor  a  graphic  solution  of  (1).  This 
method  is  frequently  preferable  to  the  first  one  mentioned 
above. 

Such  methods  may  be  useful  in  various  ways.  If  no 
high  degree  of  approximation  is  required,  the  graphical 
result  may  be  sufficient  in  itself;  it  may  be  used  as  a 
rough  check  on  a  more  accurate  result  obtained  in  some 
other  way  ;  it  gives  a  first  approximation  that  may  be 
needed  as  a  starting  point  for  more  elaborate  methods, 
or  it  may  suggest  some  value  of  the  variable  which  by 
substitution  is  found  to  satisfy  the  equation  exactly. 

EXERCISES 

Solve  the  following  equations  graphically. 

1.    a;4- 3a-3  + 3=0.  2.    Sa:*  -  2^3  -  21  x^  -  4x  +  11  =  0. 

3.    X  +  10"^  =  0.  4.    a:  +  2  cos  x  =  0. 

5.    X  -\-  logio  X  =  0.  Q.   X  +  cos  X  =  I. 

7.  Trace  the  curve  y  =  x  sin  x,  locating  maxima  and  minima 
graphically. 

8.  Solve  the  equation  x  log  x  =  1. 

9.  A  gutter  whose  cross-section  is  an  arc  of  a  circle  is  to  be  made 
by  bending  into  shape  a  strip  of  tin  of  width  8  inches.  Find  the 
radius  of  the  cross-section  when  the  carrying  capacity  of  the  gutter 
is  a  maximum.  Ans.  2.55  in. 

,71.    The  cycloid.      In  the  remainder  of  this  chapter  we 

consider  several  special  tran- 
scendental curves. 

The  path  traced  by  any  point 
A  on  the  rim  of  a  wheel  that  rolls 
without  slipping  along  a  straight 
track  is  called  a  cycloid. 

Let  the  circle  of  radius  a  roll 
along  the  a:-axis,  and  take  the 
initial  jiosition  of  A  as  origin. 
Then,  if  (a:,  ^)  are  the  coordinates  of  A^ 


CURVE  TRACING  IN  CARTESIAN  COORDINATES  "^l^ 
Y 


Fig.  48 


OB  =  arc  AB  =  aO, 
x=  OB-  AC=ae  -a^m  0  =^  a{6  -  sin  6), 
y  =  BO'  -  CO'  =  a-aGosd^  a{l  -  cos  6). 

These  are  the  parametric  equations  of  the  cycloid  in  terms 
of  the  angle  6  through  which  the  circle  has  rolled.  The 
coordinates  of  the  center  of  the  rolling  circle  are  (a^,  a). 
The  rate  at  which  the  center  is  advancing  is 

d  r    n\  du 

at  at 

where  o)  is  the  angular  velocity. 
The  curve  is  shown  in  Fig.  48. 

72.  The  epicycloid.  If  a  circle  rolls  without  slipping 
on  the  outside  of  a  fixed 
circle,  a  point  on  the 
circumference  of  the  roll- 
ing circle  generates  an 
epicycloid. 

Let  a  and  h  be  the 
radius  of  the  fixed  circle 
and  that  of  the  rolling 
circle  respectively,  and 
suppose  the  point  A  was 
originally  at  E.  Then 
arc  AL  =  a»rc  UL, 


or 


hcj)  =  ad. 


I    Fig.  49 


The  equations  of  the  path  of  A  in  terms  of  the  parara- 


^ 


CALCULUS 


eter  6  may  be  obtained  as  follows : 

X  =  0M=  OF+FM=  0F+  DA 

=  (a  +  5)  cos  6  -\-h  ^\n 


*-(!-. jj 


=  (a  +  6)  COS  ^  —  5  cos  {6  +  <^) 
=  (a  +  6)  cos  6>  -  5  cos  (<9  + 1^) 

=  (a  4-  5)  cos  6  —  h  cos      ,     ^  ; 


y  =  MA  =  FI)  =  FO'  - 

=  (a  +  6)  sin  ^  —  6  cos 


h 
DO' 


(9 


=  (a +5)  sin  (9 -5  sin  ^4^(9. 

"~73.  The  hypocycloid.  A  point  on  the  circumference 
of  a  circle  that  rolls  on  the  inside  of  a  fixed  circle  gener- 
ates a  hypocycloid. 

Its  equations  are  obtained  in  the  same  manner  as  those 
of  the  epicycloid.     They  are 

x  =  (a  —  h^  cos  6  -\-h  cos  — - 

a  —  h 


0, 


y  =  (^a  —  J)  sin  ^  —  J  sin  '^^ — 6. 


EXERCISES 

1.  Show  that  the  tangent  to  the  cycloid  passes  through  the  highest 
point  of  the  rolling  circle. 

2.  A  wheel  of  radius  2  ft.  rolls  on  a  straight  track  with  a  velocity 

of  6  ft.  per  second.     Find  jx,  jy,  j,  jti  and _/„  at  the  points  ^  =  0,  ^  =  ^, 

^  =  TT. 

3.  The  highest  point  on  an  arch  of  the  cycloid  js  called  its  vertex. 
Show  that  by  taking  the  origin  at  the  vertex  and  replacing  6  by 
01  —  0  —  TT,  the  equations  of  the  cycloid  become 

x'  =  a(^'  +  sin^'), 
y  =-  a(l-cos^'), 


CURVE  TRACING  IN  CARTESIAN  COORDINATES     Hi 

or,  if  we  change  the  sense  of  the  ?/-axis  and  drop  subscripts, 

X  —  a(d  -\-  sin^), 
y  =  a(l  —  cosO). 

4.  Sketch  the  epicycloid  for  which  the  rolling  circle  and  the  fixed 

circle  have  the  same  radius.     If  -—  =  —  radians  per  second,  find  v  and 

at       2 

J,  and  also  find  between  what  limits  these  quantities  will  vary. 

5.  Show  that  the  hypocycloid  for  which  b  =  -r  is  a  diameter  of  the 
fixed  circle. 

6.  Show  that  the  equations  of  the  hypocycloid  of  four  cusps,  for 

which  b  =  -,  may  be  written 

X  =  a  cos^  6, 

y  =  a  sin^  0. 
Hence  find  its  cartesian  equation.     Trace  the  curve. 

7.  Give  the  cartesian  equation  of  the  cycloid. 

8.  Find  the  radius  of  curvature  of  the  cycloid.         Ans.  4  a  sin  -. 

9.  Trace  the  epicycloid  for  which  &  =  -  . 


CHAPTER   X 


CURVE  TRACING  IN  POLAR  COORDINATES 

74.    Slope   of   a  curve   in  polar  coordinates.     We   have 
seen  that  in  sketching  a  curve  it  is  helpful  to  know  the 

direction  of  the  curve  at  any  point. 
In  cartesian  coordinates  the  direc- 
tion at  the  point  P^  :  (x^^  y^  is  most 
easily  determined  by  giving  the  in- 
clination of  the  curve  to  the  straight 
line  y  =  y^  —  i.e.  the  "slope"  of 
the  curve  —  since  this  is  found  by 
a  mere  differentiation.  Similarly, 
given  the  equation  of  a  curve  in 
polar  coordinates,  the  direction  at 
the  point  Pq  :  (r^,  0^  is  best  found 


Fig.  50 


by  means  of  the  inclination  to  the  curve  r  =  r^, 
of  course  a  circle  through  P^  with  center  at  0. 
reason  the  quantity  tan  (/>,  where  <^ 
is  the  angle  between  the  curve  and 
the    circle    just   mentioned,  will   be 
called  the  polar  slope. 

To  find  the  polar  slope  we  proceed 
as  follows.'  Consider  a  fixed  point 
P  :  (r,  ^)  and  a  neighboring  point 
P'  :  (r  4-  Ar,  6  +  A^)  on  the  curve 
(Fig.  51),  and  drop  a  perpendicu- 
lar PJSr  from  P  upon  OP'.  Let 
<i>'  =  ANPP'.     Then 

NP'       OP'  - 


which  is 
For  this 


Fig.  51 


tan  (^'  = 


ON 


pjsr 


pjsr 


112 


CURVE  TRACING  IN  POLAR  COORDINATES^^ 


But 


OP'  =r  +  Ar,  0N=  r  cos  A6>,  PN=  r  sin  A(9. 


Hence 


tan  </>' 


r  +  Ar  —  r  cos  A^ 
r  sin  A^ 

Ar  +  r(l  —  cos  A^) 
r  sin  A^ 

Ar  +  2  r  sin^  \  A(9 
r  sin  A^ 

Ar   ,  r  sin  i  A^    .     i   *  zi 
A6>  J  A6>  2 


sin 


A(9 


I 


When  P'  approaches  P  along  the  curve,  </>'  approaches 
(/).      By  §  38, 

sin  a 


Hence 


lim 
a->o     a 


=  1. 


tan  </)  =  lim  tan  (f>'  = 


dr 


A0->O 


or 


4.         JL         ^1" 

tan  9  =  — -. 

The  formula  for  tan  cj)  may  be 
remembered    as     follows.      Strike 
through  P  a  circular  arc  PM  with 
center  at  0  (Fig.  52).     Then 
arc  PM  =  rdO, 

MP'  =  dr  (approximately), 
and 

MP' 


tan  (f)  = 


(approximately  ) . 


arc  PM 

This  at  once  suggests  the  formula. 
I 


Fig.  52 


114  CALCULUS 

cliT 

75.    Maxima  and  minima.     When  -—  =  0,  <^  =  0,  and  in 

du 

general  r  is  a  maximum  or  a  minimum,  as  at  A  and  B^ 
Fig.  53.  Just  as  in  cartesian  coordinates,  there  is  a  pos- 
sible exception  :  the  curve  may  have 
the  form  shown  at  C.  However,  the 
exceptional  case  is  rare  in  the  simpler 
curves. 

76.    Curve  tracing.    Before  sketch- 
ing a  curve  whose  equation  is  given 
in   polar  coordinates,  the  following 
questions  should  be  considered*  : 
(V)  Is   the    curve    symmetric   with 
0  respect  to  the  initial  line  ?     (It  is,  if 

the  equation  is  unchanged  when  6  is 
replaced  by  —  ^;   other  tests  may  frequently  be  used.) 

(2)  Is  it  possible  to  determifie  any  particular  regions  of 
the  plane  within  which  the  curve  must  lie  ? 

(3)  At  what  poiiits  is  the  polar  slope  0  ?  Is  the  radius 
vector  a  maximum  or  a  minimum  at  each  of  these  points  f 

The  above  discussion  is  frequently  insufficient  to  deter- 
mine the  general  form  of  the  curve,  in  which  case  addi- 
tional points  must  be  plotted. 

Example :     Trace  the  lemniscate  r^  =  cl^  cos  2  6, 

(1)  The  curve  is  symmetric  about  the  initial  line. 

(2)  cos  2  ^  is  negative,  and  r  is  imaginary,  when 


also  when 


—  <  At)  <  -—,  I.e.  -T  <  c/  <  — -; 
2  2  4  4 


A  2  4  4 


Hence  the  curve   lies  entirely  within  the  sectors  AOB^ 
COD  (Fig.  54). 

*  For  brevity,  the   discussion  of  singular  points   and  asymptotes  is 
omitted. 


CURVE  TRACING  IN  POLAR  COORDINATES     115 
(3)  Since 


r  dr  =  —  a?  sin  2  6  dO^ 

-  a^  sin  2  e 


tan  (^  = 


0^  cos  2  ^ 


=  -  tan  2  e. 


From  this  the  direction  of  the  curve  at  any  point  may  be 
found. 
When 

tan  (^  =  -  tan  2  ^  =  0, 

2  ^  =  0,    TT,    2  TT,    3  TT, 

and 

2  2 

Only  the  values  0  and  it  give 
real  values  of  r  ; '  at  each  of  these 
points  r  is  a  maximum,  viz.  r  =  a. 
The  curve  passes  through  the 
origin     whenever     r  =  0;     i.e. 

when  cos  2  ^  =  0,  or  ^  =  — ,  — — , 

4      4 

etc.      The  curve  is  shown  in  Fig.  54. 

EXERCISES 

Trace  the  following  curves. 

I.  r  =  2  a  cos  d.  2.    The  spiral  of  Archimedes  r  =  aB. 
3.    r  =  a  sec  ^.  4.    r  =  a  cos  2  6. 

5.    r^  =  a^  sin  6.  6.    r  =  a  cos  3  ^. 

7.  r2  sin  2  ^  =  a^. 

8.  The   linia^on   r=ft  — acos^,    (a)  when   h  =  2a\     (h)  when 
J  =  a  ;   (c)  when  h  =  \a. 

9.  The  conic  r  = 7;,  where  e  is  the  eccentricity,  (a)  if 

1  —  e  cos^ 

e<l;  (&)  if  e  =  l;   (c)  if  e>l. 

10.    The  logarithmic  spiral  r  =  e*^.     Show  that  tan  <^  is  constant. 

II.  What  is  the  form  of  the  curve  r  =  a  cos  nO,  (a)   when  n  is 
even;  (6)   when  n  is  odd? 


CHAPTER   XI 
THE   INDEFINITE   INTEGRAL 

77.  Integration.  We  have  been  occupied  up  to  this 
point  with  the  problem  :  Given  a  function,  to  find  Its  de- 
rivative. Many  of  the  most  important  applications  of  the 
calculus  lead  to  the  inverse  problem  :  Given  the  derivative 
of  a  function,  to  find  the  function.  The  required  function 
is  called  an  integral  of  the  given  derivative,  or  integrand^ 
and  the  process  of  finding  it  is  called  integration. 

If  f(x)  is  a  given  function  and  F(x')  is  a  function 
whose  derivative  is  f(x)^  the  relation  between  them  is  ex- 
pressed by  writing 


F(x)  =  j'  f{x)dx. 


where  the  "  integral  sign  "  j  indicates  that  we  are  to  per- 
form the  operation  of  integration  upon  f(x)dx.  For 
reasons  that  will  appear  later,  it  is  customary  to  write 
after  the  integral  sign  the  differential  f(x)  dx,  rather  than 
the  derivative /(a:). 

Examples:  (a)  Find  the  equation  of  a  curve  whose 
slope  at  every  point  is  equal  to  twice  the  abscissa  of  the 
point. 

We  have  to  find  a  function  y  such  that 

-^  =  2  x^  or  dy  —  2  X  dx  ; 
dx 

hence 

7/  =  \  2  X  dx. 
116 


THE  INDEFINITE  INTEGRAL 


117 


It  appears  at  once  that  2  a;  is  the  derivative  of  ^.     Thus  a 
curve  having  the  desired  property  is  the  parabola 

y  =  x^. 
But  it  is  clear  that  if 
(1)  y  =  :r2  +  0,        • 

where  0  is  any  constant  whatever,  we  still  have 

dy  =  2  X  dx^ 

and  our  data  are  satisfied  by  any 
one  of  the  family  of  parabolas  rep- 
resented by  (1).  In  order  to  obtain 
a  unique  answer  to  our  problem,  we 
must  have  some  additional  informa- 
tion about  the  curve.  Thus,  if  it  is 
to  pass  through  the  point  P :  (1,  |), 
we   substitute   these   coordinates   in 

(1)  = 

and  the  answer  is 

y  =  x^-\-\. 

(b)  Find  the  velocity  of  a  body  falling  freely  under 
gravity  at  the  end  of  5  seconds,  if  the  initial  velocity  is 
20  ft.  per  second  upward. 

Taking  motion  downward  as  positive,  we  have  to  find  v 
from  the  relation  (see  §  55) 


Fig.  55 


dv 


Hence 


V  =  i  g  dt  =  gt  -\-  C. 

Making  use  of  the  fact  that  v=  —  20  when  ^  =  0,  we  get 

-  20  =  0  +  C, 
and  the  velocity  at  the  time  t  is 

i;=^^-20. 


118  CALCULUS 

At  the  end  of  5  seconds  we  have,  taking  g  =  32, 

V  =  140  ft.  per  second. 

(f?)  Find  the  space  covered  in  the  fifth  second  of  the 
motion  in  (5). 
Here 

v=—=S2t-20, 
dt 

X  =  r(32  t  -  20}dt  =16t^-20ti-O. 

No  data  are  given  for  determining  O.  But  if  we  denote 
by  Xt  the  space  covered  in  t  seconds,  the  space  described 
in  the  fifth  second  is 

■  2^5  -  2:4  =  (16  .  25  -  20  .  5  +  (7)  -  (16  .  16  -  20  .  4  +  (7) 
=  124  ft., 

the  unknown  constant  having  disappeared.  In  fact,  0  is 
merely  Xq,  the  distance  of  tlie  starting  point  from  some 
arbitrarily  chosen  origin,  so  that  the  distance  passed  over 
between  any  two  instants  must  necessarily  be  independent 

of  a 

78.  Integration  an  indirect  process.  Differentiation  is  a 
direct  process ;  by  means  of  the  fundamental  formulas 
the  derivative  of  any  elementary  function  may  be  found. 
On  the  other  hand,  to  find  an  integral  of  a  given  function, 
we  must  be  able  to  discover  a  function  whose  derivative 
is  the  given  integrand,  and  this  is  always  in  the  last 
analysis  a  matter  of  trial.  The  problem  can  by  no  means 
always  be  solved  ;  in  fact,  there  are  many  comparatively 
simple  functions  whose  integrals  cannot  be  expressed  in 
terms  of  elementary  functions*. 

79.  Constant  of  integration.  In  each  of  the  examples 
of  §  77,  an  arbitrary  constant  presented  itself.  It  is  clear 
that  this  will  be  the  case  in  general ;  i.e.  a  function  whose 

*  It  will  be  shown  in  §  81  that  for  every  continuous  function  an  integral 
exists^  although  it  may  not  be  an  elementary  function. 


THE  INDEFINITE  INTEGRAL  119 

derivative  is  given  is  not  completely  determined,  since  it 
contains  an  arbitrary  additive  constant,  the  constant  of 
integration. 

On  account  of  the  presence  of  this  undetermined  con- 
stant, the  function  if(x)dx  is  called  the  indefinite  integral 
oifix). 

80.  Functions  having  the  same  derivative.  In  §  79  it 
was  tacitly  assumed  that  if  the  derivative  of  a  function  is 
given,  the  function  is  determined  aside  from  an  additive 
constant.      That  this  is  true  follows  from  the 

Theorem  :  Two  functions  having  the  same  derivative 
differ  only  hy  a  constant. 

The  theorem  is  almost  self-evident.  Let  <i>(x)  and  ^(x) 
be  the  two  functions,  and  place 

y  =  <i>ix)-^^r{x). 
By  hypothesis, 

dx 

The  rate  of  change  of  y  with  respect  to  x  is  everywhere  0, 
hence  y  is  constant. 

EXERCISES 

Evaluate  the  following  integrals,  checking  the  answer  in  each  case 
by  differentiation. 


dx. 


1.  (rt)^^;^^;                {h)^(2x-x'^)dx]  (c)  f  (1  -  4/4)r7f ; 
(r/)|(l  +  3/)%;      (6)0;  (/)|(^^--^) 

2.  J^.  3.    ^^inOdd.  4.     rsin2^^^. 
5.     (  Vx  +  1  dx.  6.     j  Vl  -  X  dx.  7.     i  e'^dx. 

8.     ({l-\-2xydx.    9.    JxVa^  +  x^  dx.         Ans.  ^(a'^  +  x^y  +  C. 

10.  Find  the  equation  of  the  family  of  curves  whose  slope  at  every 
point  is  equal  to  the  square  of  the  abscissa  of  the  point.  Exhibit 
graphically. 


120 


CALCULUS 


11.   Find  the  equation  of  that  one  of  the  curves  of  Ex.  10  that 


Ans.  y  =  —  —  14. 
^       3 


passes  through  the  point  (8,  —  5). 

12.  A  body  falls  from  rest  under  gravity.  Find  the  velocity  at 
the  end  of  3  seconds,  and  the  distance  traveled  in  that  time. 

13.  Find  the  equation  of  the  curve  for  which  y"  =  4  at  every 
point,  if  the  curve  touches  the  line  ^/  =  3  x  at  (2,  6). 

Ans.  y  =  2  x'^  —  5  X  -\-  8. 

14.  A  body  moves  under  an  acceleration  numerically  equal  to  the 
time.  If  the  initial  velocity  is  10  ft.  per  second  in  the  direction  of  the 
acceleration,  find  v  and  x  at  the  end  of  4  seconds,  x  being  measured 
from  the  starting  point. 

15.  In  Ex.  14,  find  the  initial  velocity  if  the  body  moves  10  ft.  in 
the  first  second. 

16.  Find  the  equation  of  the  curve  for  which    y"  = ,  if  the 

x'^ 
curve  makes  an  angle  of  45°  with  OX  at  the  point  (1,  0). 

17.  Find  the    equation    of   the   curve    through    (1,  2)    and  (2,  3) 

(«)  if  y"  =  0;  (b)  if  y"  =  Q  x ;   (c)  if  y"  =—.     Trace  the  curve  in 

x^ 
each  case.  .4/^s".  (6)  y  =  x^  —  6x  +  7. 

81.  Geometric  interpretation  of  an  integral.  Consider 
the  area  A  bounded  by  the  curve  ^  ==y(a:),  the  a:-axis,  the 

fixed  ordinate  x  =  a^  and  a  vari- 
able ordinate  x  =  x.  This  area 
is  evidently  a  function  of  x. 
We  proceed  to  find  the  deriva- 
f{x-\-Ax)    ^i^®  ^f  -^  with  respect  to  x. 

When  X  is  increased  by  an  ^.^ 
2:  ^  amount  Ax^  A  assumes  an  in- 
crement  A^,  the  area  KLRN 
in  Fig.  56.  It  appears  from  the 
figure  that  AA  is  greater  than  the  area  f(x)Ax  of  the 
inscribed  rectangle  KLMN^  and  less  than  the  area 
f(x-\-  Ax^Ax  of  the  circumscribed  rectangle*: 

f(x)Ax<  AA<f(x  + Ax^Ax. 

*  The  argument  is  readily  modified  to  fit  the  case  when  f{x)  is  a  decreas- 
insr  function. 


0 


X 


Fig.  56 


THE  INDEFINITE  INTEGRAL  121 

Hence 

If  now  Ax  approaches  0,  f{x  +  A^:)  approaches  /(a:),  and 

AA 

since always  lies  between /(:r)  and/(2:  +  Aa;),  it  must 

Ax 
also  approach /(a:).     Thus 

dA      T      A^      ..  . 

—  =  lim  =j{x). 

dx         Aar->0  Ax 

Since  the  derivative  of  A  is /(a:),  it  follows  by  the  defi- 
nition of  the  integral  that 


=  jf(x)dx. 


In  case  the  position  of  the  fixed  ordinate  x  =  a  is  given, 
the  constant  of  integration  may  be  determined  by  the  fact 
that  ^  =  0  when  x  =  a. 

We  have  thus  proved  the  following  result : 

•         ^]i£^irUf/n,ifji  integral  ifCx^dx  represents  the  area  hounded 

hy  the  curve_]j  =zJls)'>  the  Xj^JOlxis,  a  fixed_ordinate,  and  a 
variabl£~M^dinate. 

It  is  evident  that  if /(a:)  is  continuous,  this  area  always 
exists ;  hence  every  contirmous  functioyi  has  an  integral 
(cf.  §  78).  " 

Since 

the  formula  for  the  area  uiider  the  curve  is  frequently 
written  r/r{y  \ 

ly  A=  \y^^ 

Example:     Find    the    area    bounded    by  the    parabola 
y  =  x^^  the  a^-axis,  and  the  lines  x  =  1^  x  =  4. 
The  area  from  a;  =  1  to  any  variable  ordinate  is 


A=  \  ydx=  \  x^dx=  —  -{-  O. 


122  CALCULUS 

Since  A  =  0  when  a:  =  1,  we  have 

0  =  3^4-C,    C  =  —  3, 
or 

"^-3      3 
In  particular,  the  area  from  a:  =  1  to  a;  =  4  is 

EXERCISES 

In  the  following,  find  the  area  bounded  by  the  x-axis,  the  given 
curve,  and  the  indicated  ordinates.  Check  roughly  by  drawing  the 
figure  on  coordinate  paper  and  estimating  the  area. 

1.  y  =  x^,   X  =  0,   X  =4: .  Ans.   64. 

2.  The  parabola  y^  =  4i  x  and  its  latus  rectum. 

3.  The  hyperbola  y  =-,   x  =  1,   x  =  3.  Ans.   1.099. 

X 

4.  Find  the  area  of  one  arch  of  the  sine  curve.  Ans.   2. 

5.  Find  the  area  bounded  by  the  parabola  y  =  1  —  x^  and  the  x-axis. 

82.  Variable  of  integration.  In  the  last  article  we  had 
occasion  to  use  the  symbol    j  ?/  dx.    In  order  that  such  a 

symbol  shall  have  any  meaning,  j/  must  be  directly  or 
indirectly  a  function  of  x.  The  variable  whose  differential 
occurs  is  called  the  variable  of  integration^  any  other 
variables  appearing  under  the  integral  sign  must  be 
functions  of  the  variable  of  integration,  and  their  values 
in  terms  of  that  variable  must  be  introduced  before  the 
integral  can  be  evaluated. 

Xhaiact  that  the  differential  occurring  tells  us  which 
variable  is  the  variable  of  integration  is  one  of  the  reasons 

for  using  the  notation    i  f(x)dx  rather  than  the  notation 

J/(^). 

83.  Change  of  the  variable  of  integration.  If  x  is  so  re- 
lated to  y  that 

(1)  (f>(x)dx  =  y\r(^y)dy, 


THE  INDEFINITE  INTEGRAL  123 

we  may  replace   j  (j>(x)dx  by  j  "^(^y^dy.     For,  let 
^(x)=   C(f)(x)dx, 

^(^)  =  ffOj)dy. 
Now  ^ 

r:;?2;  c?^  c^a;  dx 

The  two  functions  therefore  have  the  same  derivative,  by 
(1),  and  hence  differ  only  by  a  constant.  Since  each 
function  contains  a  constant  of  integration,  these  constants 
inay_be„so_  chosen  that 

The  device  of  replacing  a  given  integral  by  an  equiva-   ^  1 
lent  integral  in  a  different  variable  is  very  useful  in  many 
problems. 

84.    Integration  by  substitution.     A  change  of  variable 
is  usually  brought  about  by  means  of  an  explicit  substitu-    ^^"' 
tion 

j^^^^dx  =  (^'  (%i)du. 

The  process  is  called  integration  hy  substitution^  and  is 
"^ighly'' important.  It  is  tu  be  remembered  that  not  merely 
x^  hut  dx  as  well,  must  be  replaced  by  the  proper  value  in 
terms  of  the  new  variable. 

Example  :     Evaluate  \  x^l  —  x  dx. 

Let  us  put 

1  —  a;  =  w, 
so  that 

a:  =  1  —  u,  dx  =  —  du. 


124  CALCULUS 

Then 

j  xVl  —  xdx=—  1(1  —  u)u^ du=  —  j  (u^  —  u^'ydu 

=  - 1(1 -2:)^ +  1(1 -2:)^+ a 

EXERCISES 

1.  Work  the  above  example  by  placing  I  —  x  =  u\ 

2.  Evaluate  i  x(3  —  2  x^)dx  by  substituting  S  —  2  x^  =  u.     Check 
by  expanding  the  given  integrand  and  integrating  directly. 

Evaluate  the  following  integrals  by  means  of  the  indicated  sub- 
stitution, and  check  the  results  by  differentiation. 

3.  I  Va  +  bx  dx,  a  +  hx  =  u. 

4.  I  sin^  6  cos  0  dO,  sin  6  =  u. 

5.  \z -: — ;,  2  X  =  U. 


h 


+  4a:;2 


6.    CJ^A^,  1-x^  =  u. 


n    ~Z — Z'  l  +  tan^  = 

•^  1  +  tan  6 

8,  J 


7 

J  1 


u. 


Vl  -  x^ 

9.  If  the  velocity  of  a  point  moving  in  a  straight  line  is  given  as 
a  function  of  the  time,  show  that  the  distance  covered  may  be  found 
by  the  formula 


=J^ 


dt. 


Given  v  =  10  t  +  20,  find  x  in  terms  of  t  by  substituting  the  value  of 
V  in  the  above  integral ;  also  find  x  in  terms  of  v  by  substituting  for 
dt.     Show  that  the  two  values  of  x  are  equivalent. 

10.    Given  x  =  t^,   y  =  dt,  find 

A  =  \  y  dx 

as  a  function  of  t  by  substituting  for  y  and  dx  their  values  in  terms 
of  ^  Ans.   A  =:2t^  +  C. 


THE  INDEFINITE  INTEGRAL  125 

11.  In  Ex.  10,  find  y  in  terms  of  x  by  eliminating  t.     Then  express 
yl  as  a  function  of  x  by  substituting  for  y  and  integrating  with  re- 

3 

spect  to  X.  Ans.   A  =  2  x^  +  C. 

12.  In  Ex.  11,  find  A  as  a  function  of  y  by  substituting  for  dx. 

Ans.    A  =  ijy^  +  C. 

13.  Show   that    the    three  values  of   A  found  in  Exs.   10-12  are 
equivalent. 

14.  By  the  formula 

A  =  j  ?/  dx, 

find  the  area  under  the  curve  y'^  =  x  from  x  =  0  to  x  =  4,  («)  by  sub- 
stituting for  y  in  terras  of  x ;  (&)  by  substituting  for  dx  in  terms  of 
y  and  dy. 

15.  Proceed  as  in  Ex.  14  for  the  area  under  the  curve  y  =  e^  from 
a;  =  0  to  a:  =  1. 

16.  Proceed  as  in  Ex.  14  for  the  area  of  half  an  arch  of  the  curve 
y  =  \  sin  4  x.  Ans.  x\. 


CHAPTER   XII 
STANDARD  FORMULAS  OF  INTEGRATION 

85.  Standard  formulas.  If  we  were  to  try  at  present  to 
solve  any  but  the  simplest  applied  problems  involving 
integration,  we  must  fail  through  inability  to  evaluate 
the  indefinite  integrals  involved.  We  shall  therefore 
devote  this  and  the  following  chapter  to  the  technique  of 
integration  —  the  formal  evaluation  of  indefinite  integrals 
—  after  which  the  question  of  applications  will  be  treated 
at  some  length. 

As  the  first  step  toward  facility  in  integration,  the 
student  must  become  thoroughly  familiar  with  the  fol- 
lowing 

Fundamental  Integration  Formulas 

(1)  fdu=u  +  C. 

(2)  f(du  -\-  dv)  =fdu  +  Cdv, 

(3)  Ccdu  =  cCdu, 

/i,n+l 
n-hl 

(5)  f^  =  logu-\-Cr\ 

(6)  fe^dw  =  e"  +  C, 

(7)  J  cos  w  i/z/  =  sin  M  +  C  =  cos  f  M  —  ^  j  +  C, 


126 


STANDARD  FORMULAS  OF  INTEGRATION       127 

(8)  \  sin  u du  =  —  cos  u  -\-  C  =  sin  I u  —  -]-{-  C, 

(9)  j  sec^  udu  =  tan  w  +  C, 

(10)  /. 


du  .    u  ,  ^ 

=  arcsin  -  +  C, 


Va^  -  u^  a 


<-'">  S^ 


du         1       4.«    w  ,  ^ 
=  -  arctan  -  +  C, 


+  w^      a 


(12)    j  udv=uv—  i  vdu. 


It  is  strongly  recommended  that  each  of  the  formulas 
be  written  out  by  the  student  in  words,  and  memorized  in 
that  form. 

The  test  of  the  correctness  of  an  iiitegral  is  that  its 
derivative  must  Jje  Jhe^^iven^  jntegrand.  The  above 
formulas  may  be  verified  at  once  by  differentiation. 

86.  Formulas  (l)-(3).     Formula  (1)  merely  embodies 
'the  definition  of  an  integral. 

Formula  (2)  is  readily  extended  to  the  case  of  any 
number  of  terms.  The  formula  shows  that  if  the  inte- 
grand consists  of  a  sum  of  terms  each  term  may  be  inte- 
grated separately. 

Formula  (3)  says  that  if  the  integrand  contains  a  con- 
stant factor,  that  factor  may  be  written  before  the  integral 
sign.  As  a  corollary,  we  may  introduce  a  constant  facto?' 
into  the  integrand^  provided  we  place  its  reciprocal  before 
the  integral  sign.  But  the  student  must  beware  of  intro- 
ducing variable  factors  by  this  rule. 

87.  Formula  (4) :  Powers.  This  formula  evidently 
fails  when  n  =  —  l.  The  exceptional  case  I —  is  taken 
care  of  by  formula  (5). 


128  CALCULUS 

Exaynples : 

(a)     CU  2;3  4-  1  4-  -^^dx  =  8j"a;3  dx  -\-Jdx  +  If 3^''^  dx 


=  — —  -A-x  —  -X  ^  +  C' 
4  2 

4  2x 


(b)  Jx(l  +  :r2)2  dx  =f(x  +2x^-\-  a^)dx 


/VIM  ry*^  /y»0 


A  better  method :     Introducing  a    factor    2    after  the 
integral  sign  and  its  reciprocal  in  front,  we  have 

fx(l  +  x^)^dx  =1/(1  +  x'^y  -2xdx. 

Since  2  x  dx  is  the  differential  of  1  -\-  x^^  formula  (4) 
applies  with  u=l  -\-  x^^  n  —  2.     Hence 

i  fci  +  x^y  .2xdx  =  ^il±^  +  6^=  1(1  +  x'^y  4-  C. 
2^  2        3  t) 

In  substance  we  have  introduced  a  new  variable  u  =4  i^^ 
as  in  §  84.  But  with  a  little  practice  one  is  able  to  think 
of  the  quantity  1  -\-  x^  directly  as  the  variable  of  integra- 
tion, without  writing  out  a  formal  substitution,  thus 
effecting  a  great  saving  of  time. 

The  student  should  compare  the  two  answers  that  have 
been  obtained  in  this  example. 

Va  -j-  bx  dx  =  -  1  Va  -\-hx  -  h  dx 
_  1  {a-\-  hx)  2      ^ 


O  0 

Here  the  quantity  a  -{-  hx  is  taken  as  the  variable  of  inte- 
gration, the  factor  h  is  introduced  to  give  the  proper 
differential,  and  (4)  then  applies  with  n=  |. 


STANDARD  FORMULAS  OF  INTEGRATION      129 

EXERCISES 

Evaluate  the  following  integrals;  check  the  results  by  differen- 
tiation. 

1-    1(1+-,)''^-  2.    j"(v';+2V^  +  -i^)rfx. 

'    3.    j'-£i+ji±-lrf,,  4.    j4(:.+  l)Vix. 

6.    ^(1  -  2  xyiJx.  6.     ({X  +  \){x  +  2)  dx. 

9.     f  x\/«2  +  a--^/.r.  10.     f--^'^^-^    • 

c/  ^  Vl  -  ^3 

11.     r(«2_a;2)2f/x.  12.     f  V(I+17)=^r/^ 

13.     fe^vTT^'/y.  14.     f(x- +  l)(x2  + 2a:+ 6)r/a;. 

J         2  2      3    -/,.  2  2      5 

16.     rsin2  6lcos^^/^.  17.     r     ^^"^-^     • 

18.     r(rt' -xh^f/x.  19.     f  —  ^^^       • 

*^  -^  V;^  -  4  X 

20.     (aVflj-vix.  21.     I — ~ 

V  "^  (1  -  0 

8&N.,Formulas    (5)-(6)  :    Logarithms   and   exponentials. 

— • 

a^  -f  ar 

If  the  factor  2  be  inserted  under  the  integral  sign,  the 
numerator  becomes  the  differential  of  the  denominator, 
and  (5)  applies : 

—  dx. 

x  +  1 

K 


130  CALCULUS 


By  division  we  find 

=  X—  z  -] 


x+  1  x-[-l 

Whence 

f^^dx=  f(x-2-^^-)dx 
^  x-\-l  ^  \  x  +  lj 


x^ 


(<?)   Evaluate    I  e^""  dx. 

If  the  factor  3  be  inserted,  this  is  integrable  by  (6)  ; 
J'eS-  dx  =  ije^-  -^dx^^e^^^  C. 

In  (5)  we  have  used  a  device  that  finds  frequent 
application : 

-4s  the  first  step  toward  integrating  a  rational  fraction^ 
carry  out  the  indicated  division  until  the  numerator  is  of 
lower  degree  than  the  denominator. 

EXERCISES 

Evaluate  the  following  integrals ;  check  the  results. 

1.    j"-^-  2.  j"^^^- 

-^  X  -7  J  VI  -  4a: 

3      f-^^.  4.    f-^^- 

^'    J  1  -ix  J 


7. 


VI  +  X 


8.  f  tan  0^^.  ^ws.     -  log  cos  6  +  C. 

9.  (ue^'-^du.  10.     P^  ~  ^^x. 


/    5  _?\ 2 

11.    JV^'-^"V    c/x.  12.    I 


ax. 


13. 


STANDARD  FORMULAS  OF  INTEGRATION       131 

x'^  +  2  a;  —  1  ,^  i  J      fsin  x  dx 


J  \  +  x'  J 


cos^a: 


15.      r_S^^Mi^.  16.      P'-'^^+^/g;. 

J  VI  +  tan  <^  -^       3x  -4: 

17  r^  18  r  e'^dx 

'    J  e^'  '    ^  (1  -  e-^^)2* 

19.    Find  the  equation  of  a  curve  through  (1,  1),  if  the  slope  at 
every  point   is   inversely  proportional    to   the  abscissa  of   the  point 

(;...i£y=^). 
89.    Formulas  (7)-(9):  Trigonometric  functions. 

EXERCISES 

Evaluate  the  following  integrals. 
1.    (^m2xdx.  2.    fsec2^^^. 


3.     (x&inx'^dx.                                       4.    ^ cos  ktdt. 
5.     I  cos  (log  a:)  —  •  ^-     ) —^^ 


7.     rtan2^r70.  .4  ns.     tan  ^  -  ^  +  C. 

sin  2  X  dx 
+  h  cos  2  x 


g      C   Hm2xdx    ^  9_     Ce^cose'dx. 

J  a  -[•  h  cos  2  X  -^ 


10.  f  sin3  u  du.  Ayis.     -  cos  u  -{-  ^  cos^  w  +  C 

11.  f ^^^.  12.     C  sin  0  cos  edO. 

J  1  +  sin  (9  ^ 

13.     (e""^  *  sin  a;  dx.  14.     f  tan  3  ^of^ 

15.    Jsec^.^^.  16.    J«i"^^^ 


COS" 


e 


17.    If  the  acceleration  of  a  moving  particle  is 

- —  =  —  k^  cos  kt, 

dfi 

the  particle  has  simple  harmonic  motion  (§  229).  Find  v  and  x  in 
terms  of  t  \i  v  =2  0  and  x  -I  when  f  =  0.  Show  that  x  vanishes 
periodically,  hence  the  particle  oscillates  about  the  origin.  Find  the 
maximum  velocity. 


132  CALCULUS 

\ 

90.  Formulas  (lO)-(ll) :  Inverse  trigonometric  functions. 

Example  : 

/dx  C         dx  \        ,       X  -\-  1   ,    ^ 

EXERCISES 

Evaluate  the  following  integrals. 
^      C       dx       •  r     xdx 

*^  V9  -  x^  J  Vl  -  x^ 

3.  f — i£—.  Ans.     iarctan  — +C. 
-^  4:x^  +  Q  6  3 

4.  f ^^^ Ans.     Arctan  (2a:  -  1)+  C. 

^  x-^  +  4  -"  Vl  +  x2 

9.  f  ^-^'  •  10.  r_^!!^. 

'    ^  VI  -  e"^"  *^  Vl  -  e^r 

11  r ^^^ ^.  12.  f--^^^ 

•^  Va:Vl  -  X  -^  x{l  +  \og^x) 

13      r     ^^-^  14      r      3:<^-y 

91.  Formula  (12):  Integration  by  parts.  From  the  form- 
ula for  the  differential  of  a  product, 

dQuv')=udv  +  V  duy 
we  find,  integrating  both  sides, 

iiv  =  \  udv  +   IV du. 

Transposing,  we  obtain  formula  (12)  : 

i  u  dv  =  uv  —   i  V  du. 

Integration  by  this  formula  is  called  integration  hy 
parts.  The  use  of  the  formula  will  be  explained  by  the 
following 

Examples  :  (a)  Evaluate   j  xe""  dx. 


x^ 


STANDARD  FORMULAS  OF  INTEGRATION    ^-t83 
Let 


whence 


Then 


u  =  X,  dv  =  e^  dxy 
du  =  dx^  V  =  I  e'^dx  =  e^. 


I  xe^'dx  =  xe""  —  i  e'^dx  =  xe^  —  e^  -\-  O. 

(h')   Evaluate   J  log  x  dx. 

Let 

u  =  log  x^  dv  =  dx, 


whence 


J        dx 

au  =  — ,  V  =  X. 


X 

Then 

/log  xdx  =  X  log  X  —  \  X  '  —  =  X  log  X  —X  -\-  C. 

Integration  by  parts  is  highly  important,  as  it  succeeds 
in  many  cases  when  the  methods  of  the  preceding  articles 
fail.     The  success  of  the  method  depends  as  a  rule  on  our 

ability  to  choose  u  and  dv  so  that  the  integral    I  v  du  is 

easier  to  evaluate  than  the  given  one.  No  general  direc- 
tions can  be  given  for  choosing  u  and  dv  ;  if  the  new 
integral  is  no  simpler  than  the  original,  we  should  begin 
over  again  with  a  different  choice  of  u  and  dv. 

EXERCISES 

Evaluate  the  following  integrals. 
1.     yxe^'dx.  2.     f  ^  sin  t  dt. 

3.  (x^e^'dx.  Ans.   ia:V-^  -  ^ xe^'  +  ^e^*  +  C. 

^4.  i  arcsin  x  dx.  Ans.    x  arcsin  x  +  Vl  —  x^  +  C. 

5.  ix^logxdx  A71S.    Yfloga:--J+C. 

6.  i  uVl  —  u  du.  7.     \  x^  arcsin  x  dx. 


id^ 


CALCULUS 


X  tan^  X  dx.  A  ns.    x  tan  x  —  —  +  log  cos  x  -\-  C. 

Cx^Va^-  x^dx.  Ans.    -  \x^{a^  -  x'^y  -  j\(a^  -  x^y  +  C. 

(x^\/a^  -  x^dx.  11.     fcos^logsin^t/^. 

(sec^  OdO.  13.     (cos^Ode. 


8. 

9. 

I 

10. 
12. 
14. 

16.  

y  *^  Va^  —  y 


xMx  4  r      r     x^  (Zz 


J  (a2  _  ^.)f '  •    J  (1  +  :r2)-^ 

r__lML.  ^ns.    _  2/2^0,2  -  y2^i  -  |(a2  -  2/2)1  +  C. 

*^  Va2  -  ^2 

9X  Integration  by  substitution.*  An  integral  that  can- 
not be  reduced  directly  to  one  of  the  standard  forms  may 
often  be  evaluated  by  the  formal  substitution  of  a  new 
variable.  If  the  integrand  is  algebraic,  and  rational  ex- 
cept for  the  presence  of  a  single  radical,  it  may  frequently 
be  integrated  by  placing  the  radical  equal  to  a  new  vari- 
able. In  general,  if  any  simple  function  is  especially 
conspicuous  in  the  integrand,  substitution  of  a  new  vari- 
able for  that  function  is  worth  trying.  However,  no 
general  rules  can  be  laid  down  ;  skill  in  the  use  of  sub- 
stitutions comes  only  with  practice. 

/w  X  dx 
~ . 

Let 

Vx  =  w,     x  =  u\     dx  =  2u  du. 
Then 

/Vxdx  _  iy  f  u^du  _  j^  /*/-!  _      1     \  7 
1-^X    ~  "-'   1  -f-  2^2  ~  "  J  V      ~  I  +  UV 

=  2u  —  2  arctan  u-\-  O 
=  2Vx  —  2  arctan  Va;  +  C. 
'sin  X  cos  X  dx 


(6)   Evaluate  f^^ 


4-  sin  X 
*  At  this  point  the  student  should  re-read  §  84. 


STANDARD  FORMULAS  OF  INTEGRATION       135 

Let 

sin  X  =  u^  cos  xdx  =  du, 
whence 

J  sin  X  cos  X  dx  _  C  u  du  _   Cf~,  1    \  , 

1  +  sin  X         *^  1  -\-u     ^V        l  +  'w/ 
=  u  —  log  (\  -^  u)  -\-  0 
=  sin  X  —  log  (1  +  sin  x)  +  0. 

EXERCISES 

Evaluate  the  following  integrals. 

dx  ft      r    a:  +  3 


^  1  -  Vx  ^ 


VI  +  2x 


rfx. 


3      C     xdx  feSxsine^^x. 

^  (x  +  1)2  J 

5.     fa-sVa^-  a:2^/a:.  6.     (x^e^'dx. 

7.     r_£!^^£__  8.     f  sin  Vx  dx. 

J  (a-  +  x^y^'  J 

J  -^      Vl  +  tan  Q 

MISCELLANEOUS  EXERCISES 

Evaluate  the  following  integrals. 

1.  fcot^j^.  2.  r 


X  dx 


•2  _i_   o'2 


Va^  +  a- 

3.     (  ^    ~.  -^  fix  4.     i  x  cos  3  a;  <ix. 

•^2  +  3a:      '  -^ 

5.    r  ^^"-^  e     flos:  X 


dx. 


7.    y  (l  +  cos ^y  sin  ^  t/(9.  8.  (x(l  -  x^)dx. 

9-    jxe-'^'dx.  10.  rsin2  2  a:  cos  2  a:  (/x. 

11.    fi5Hi2ii^^.  12.  fi^^,.. 

•^  1  —  tan  2  ^  J  I  ^  e-x 

13.    Jfl  -  ^)'^3/.  14.  rx(i  -  x2)Va:. 

15.  r  ^^^  .  16.  r  ^'^^^  .. 

•^  (1  -  xy  J  (1-  xy 


i36  CALCULUS 

17.     \ .  18.  i  cot  X  log;  sin  x  dx. 

Jx^+d,x+20  J  ^ 

19.     f-i^.  20.     f-4??^. 

J  X  log-  X  *^  Vl  —  x® 

21.     fgSxVl  -  e^^dx.  •    22.  fcos~Jx. 

23.     Ce'^^'sec^tdt.  24  r?i±^i±-?  rix 

25.    ^(J-yh'dy-  26.  JI^+_^!>^. 

27.     f— ^-i^.  28.  f.fsec^x^/x. 

^  f  1  -  x^^  J 


29.     farctanx^/x.  30.     (^^J^^^^ 


dx. 


31.    je^^rfx.  32.    j*//(6-^)V 

rf^2^^'  34.    p-'(l  +e-)3rfx. 

35.     p(l  +^'\/x-.  36.     fsec2^tan3^^l9. 

37.    Jci  -  x3)2^/x.  38.    J(a  -  x^ dx. 

39.   Jx(«-x)t.x.  ^O-L-^ 


+  10 


41.    J-^.  42.   J.^  log(l  +  .^)rfx. 

43.     f-^^^.  44.    J-^ 

^   1    +  X8  -^   1   -   d 


e^  dx 
3e* 


45      f ^^—.  46.     f^^  +  ^^^'^'cZx. 

47.     flog^^^^-i'-  48.     f     _^^'^ 

-^  -^  Vx(l  +  x) 

49.    /.=.-"</..  60.    1^3- 

51.  Using  the  formula  cos  2  x  =  1  —  2  siii^x,  show  that 

j  sin2  jr  (f;r  =  |;i:  -  J  sin  2  a:  +  C, 

J  cos- X dx  =  ^x  +  ^sm2 X  +  C. 

52.  By  putting  x  =  a  sin  ^,  show  that 

J  Vfl^  -  jt"^  c/j:  =  2-^Va'^  -  Jf^  +  ^a^  arcsin'*^  +  C. 


i 


CHAPTER   XIII 
INTEGRATION  OF  RATIONAL  FRACTIONS 


93.  Preliminary  step.  In  dealing  with  tlie  integral  of 
a  rational  fraction,  or^iotient  of  two  polynomials,  the 
first  step  (cf.  §  88)  is  to  carry  out  the  indicated  di- 
vision until  the  numerator  is  of  lower  degree  than  the 
denominator. 
li  94.  Partial  fractions.  If  the  denominator  is  of  the 
first  degree,  or  of  the  second  degree  with  complex  roots,* 
the  integration  can  be  performed  by  the  fundamental 
formulas.  Many  examples  have  already  arisen :  e.g. 
Exs.  17,  33,  35,  40,  p.  136.  In  other  cases  the  integral 
may  sometimes  be  evaluated  either  directly  or  after  a 
suitable  substitution,  as    in  Exs.   15,  16,  27,  50,  p.   135. 

In  general,  however,  we  must  resort  to  the  method  of 
"partial  fractions."  The  first  step  (after  the  numerator 
has  been  made  of  lower  degree  than  the  denominator)  is 
to  resolve  the  denominator  into  real  linear  or  quadratic 
factors.!  I^  ^^i^  ^^^  t)e  done,  the  given  fraction  can  then 
be  expressed  as  a  sum  of  partial  fractions  whose  denomi- 
nators are  the  factors  of  the  original  denominator.  We 
proceed  to  show  how  to  do  this. 

95.  Distinct  linear  factors.  The  simplest  case  is  that 
in  which   the    denominator    can  be   broken  up  into    real 

*  That  is,  the  denominator  of  the  form  ax^  -\-  hx-\-  c  with  &'-— 4  ac  <  0. 

t  The  term  "quadratic  factor"  means  here  a  factor  of  the  second 
degree  whose  linear  factors  are  imaginary ;  i.e.  a  factor  of  the  form 
ox2  +  6x  +  c  with  62  _  4  ^c  <  0. 

137 


138  CALCULUS 

linear  factors,  none  of  which  are  repeated.     The  process 
will  be  explained  by  an 

/'x^  4-  2 
— dx. 
x^  —  X 

By  division  we  find 

rrS  +  2       -,    ,   a:  +  2 
=  j_  -f- • 

a^  —  X  7?  —  X 

The    factors    of    the    denominator    are    x^   x-\-\^    x  —  \. 
Assume 

a:+2  ^A         B  0 


■3?  —  X  X         X-\-\         X—1 

where  A,  B^  6^  are  constants  to  be  determined.     Clearing 
of  fractions,  we  find 

x  +  1  =  A(x^  -l)-^Bx(x-l)+  Cx{x  +  1). 

Since  this  relation  is  an  identity^  it  must  hold  for  all  values 

of  x.     Hence, 

putting  a;  =  0,       we  find  ^  =  —  2, 
x=-\,  B  =  l, 

x  =  h  0=^. 

Thus 

^x(x^-l}  ^\        X      2     x^\      2     x-lj 

=  x  —  2  log  X  -\-  ^  log  (a:  +  1)  4- 1  log  (a;  —  1)  -f-  O. 

EXERCISES 

Evaluate  the  following  integrals. 
1       r    ^fx  n      C    <ix 

3       Cj^jIx_^  ^^^     a;-log^^+C. 

J  x-'-4:  ^  x-2 


4. 


r(2 


2  x^  -\-  x  —  l)dx 


a;^  +  a:^  —  4  a:  —  4 
Ans.    2x+  1-1  log  {x  -  2)  -  V-  log  (a;  +  2)  +  |  log  (x  +  1)  +  C. 
I       r  xdx  g       C        x^dx 

Jl-x^'  '    J  x2  +  3  a:  +  2 ' 


INTEGRATION  OF  RATIONAL  FRACTIONS       139 

x'^  dx 

x^  -^  2  x'^  —  X  —  2 

r2  1  1  Ifi 

Ans.    -L^Ox  +  ^Jogix-  l)-ilog(x4-  l)+i^log(^  +  2)+C. 
2d  2  o 


J 


8.     icosec  6  dO.     (Note 


cosec  9  = ^  =  -■ 


1  sin  0  sin  0 


sin^      sin^^  1  —  cos^^ 

^          1  1  1  —  cos  6  ,   ^ 

2  ^  1  +  cos  ^ 

10.     \  sec  ax  dx. 


'■I 


96.  Repeated  linear  factors.  If  the  denominator  con- 
tains a  factor  (^x  —  «)%  the  above  method  fails,  since  there 
would  be  r  partial  fractions  with  denominator  a:  —  a,  and 
these  could  be  combined  into  a  single  one.  We  can  ob- 
tain the  desired  result  in  this  case  by  assuming,  corre- 
sponding to  the  factor  (^x  —  ay,  r  partial  fractions 

^    +,   ^    ,  +  -  +  ^^ 


(a;  —  ay      (2:  —  ay  ^  x  —  a 

/'    2:^—1 
-irr-^dx 
x(x  +  1.)^ 
Assume 

7^-\     _A^       B  C  D 


x{x  ^Vf~  X       {x^Yf      {x^Vy      x^-V 
0:3  _  1  =  j^(^r^  4.  1)3  j^Bx^-  Cx{x  +  1)  +  Dx{x  -I-  1)2. 
Put  2:=0:^  =  -l, 

a;  =  -  1  :  ^  =  2. 

To  find  Q  and  i>,  we  may  assign  any  two  other  values  to  x^ 
say  x  =  \  and  x=%  thus  obtaining  two  simultaneous  equa- 
tions to  solve  for  (7  and  D ;  or  we  may  equate  the  coefficients 
of  like  powers  of  x  in  the  two  members  of  the  identity. 

Equate  coefficients  oi  x^  -.    A  +  D  =  1,  D  =  2. 

Equate  coefficients  oi  x^  :    S  A -i-  0  +  21)  =  0,  O  = -1. 
Whence 

=  -log^-        -^        +-l--+21og(.r+l)+C-'. 

{X  +  1)^         X  -\-l 


140  CALCULUS 

EXERCISES 

Evaluate  the  following  integrals. 

1       r     dx  2      r      dx 

^  x^  —  x^  •'  (1  —  x)^ 

o       C     xdx  ^      C{3fi_j-_Y)dx 

■     J  (1   -  .r^5*  •     J    .rr.x-  2^2 


5 


(1  -  :c)S  J  x{x  -  2)2 


J 


13  1 

^ns.    loq:a; log  (x-\-  1) logCx  —  1) h  C. 

4      *  V    -r    y      ^      s  V  y      2(2'  - 1) 


a;4  +  12  x-3  +  52  a;2  +  96  a:  +  64 


A  -5a:— 12    ,    oi      a:4-4,   ^ 

Arts.     +  2  log — — -\-  C. 

a:2  +  6a;  +  8  ^a.-+2 


7. 


C^^ Ans.    i  e-*  -  -  +  i  log  (e^  -  2)  +  C. 


97.  Quadratic  factors.  Corresponding  to  a  factor  in  the 
denominator  of  the  form  *  ax^  +  hx  +  c  where  ^^  —  4  ac  <  0, 

we  assume  the   terms  — ;      ' ^H — 

ax'^  +  ox  -\-  c      ax^  -{•bx-{-  c 

The  case  in  which  the  denominator  has  repeated  quad- 
i-atic  factors  is  of  less  importance,  and  will  be  omitted. 

Example:     Evaluate    I —— — dx. 

Assume 
x^Jr^x  +  lQ  ^A      BC2X+2)  C 

a^  -^  2x^  -^  5x      X      x'^ -\-  2  X  -\-  f)      x^  -\-  2  x  -^  5 
2^2  -I-  4  a;  +  10  =  A(2;2  +  2  a:  +  5)  +  Bx(2  x-{-2)-{-  Cx. 

Put      2^  =  0:  5^  =  10,  ^  =  2. 

Equate  coefficients  oi  x'^ :  A  -j-  2  B  =  1,  B  =  —  ^. 
Equate  coefficients  oi  x  :  2  A-\- 2  B -^  C  =  i,  0  =  1. 

*  Of  course  such  a  factor  might  be  broken  up  into  complex  linear  fac- 
tors, after  which  the  process  of  §  95  would  apply.  The  present  method 
has  the  advantage  of  avoiding  imaginaries. 


INTEGRATION  OF  RATIONAL  FRACTIONS       141 


Whence 


J  x^  +  2 


2  +  43;+  10 


dx 


-\-zx^-\-  b  X 

=  c(^-i.  ^"+2  + I V- 

-'  \x      2     x'^-\-2x-\-  b      x^-[-  2  x-\-  0/ 
=  2  log  x  —  ~  log  (a;2  +  2  2:  -h  5)  +  -  arctan  ^-^ — h  C. 

EXERCISES 

Evaluate  the  following  integrals. 

1.     f ^'^^ ylns.  x  +  21og(a:2-4a:+5)+3arctan(a:-2)  +  C. 

-^  a:-  —  4  x  +  5 

'    J  1  +  ^4*  '    J  a:3  +  4  a;2  +  8  x' 

4.     \ A  ns.  —  log  — ■ arctan  -  +  C . 

^  o^  -  2-4  4  a  a  -  a;      2  a  a 

J.r  dx  g      r       X  r/x 

1  -  a:^  ■    J  a;"^  +  X  +  1 

•1^ 


6 

xdx 

x^-\-x'^  +  '^x  +  4 

,4  „.5.  J^  log  (a:2  +  4)  -  i  log  (x  +  1)  +  I  arctan  |  +  C. 

Q      r     x^f/a:  q      C         X  dx 

J  (1  +  x'-^)^'  ■    J  x-^  -  2  a:  +  2' 

^        x3  -^  x%l  +  x^) 

12.     C—Jl A?is.   1^+  ^log(sin(9+cos(9)+ C. 

*^  1  +  tan  6 

MISCELLANEOUS   EXERCISES 

Evaluate  the  following  integrals. 

dx 


1  r    dx  2  r^ 

*^  (a  +  bx)'^  '  ^  a  +  hx 

g      r     xdx  ^  C     X-  dx 

-^  («  +  x)2"  *  -^  (^/  +  x)2 

_      r            xdx     ■  g      r       x  dx 

J  (a  +  xy(h  +  x)'  J  (a-  +  x^yi' 


142  CALCULUS 

7.     ('-^^^d..  8.    (•        ^"        . 

^  «4  -  a;4  ^  2  a;2  +  8  a;  +  1 

1  -  x^'  '    J  {2-x) 


13 
15 


tan  6  —  cot  ^ 


f_^.  16.     f 

17.  fi^i^iy..  18  r^^  +  i  rf^. 


CHAPTER   XIV 

THE   DEFINITE   INTEGRAL 

98.  The  definite  integral.  Let /(a:)  be  a  given  function, 
F(x)  an  integral  of  f(^x')^  andjr==  a  and  x=h  two  given 
values  of  x.  The  change  in  the  value  of  the  integral  F(x) 
as  X  changes  from  a  to  6  is  called  the  definite  integral  of 
/(a;)  between  the  "-  limits  "  a  and  5,  or  simply  the  definite 
integral   from    a   to   Z>,    and   is    denoted    by    the    symbol 

Jf(jc)dx.     Its  value  is  evidently  F{h^  —  F(a^. 
a 

This  change  in  the  value  of  the  integral  between  two 
values  of  the  variable  is  required  in  many  important 
problems.  It  is  called  the  definite  integral  because  its 
value  is  independent  of  the  constant  of  integration. 

To  evaluate  a  definite  integral,  we  have  merely  to  find 
the  indefinite  integral,  and  then  subtract  its  value  at  the 
"  lower  limit  11, «  if  rom  its  value  at  the  '>  ujipe^-limit  "  h. 

It   is   custoniary  to  use    the  symbol   F(x) 
F(h)-F  (a),  „  .Thus  ^-^--.^ 

fy(x)dx  =  F(x)   '=F(b}-F(a). 

Since  the  constant  of  integration  disappears,  there  is  no 
object  in  writing  it  at  all. 

Examples :  (o^)  In  (<?),  §  77,  we  have  for  the  space  in 
the  fifth  second 

5 


a 


as  meaning 


s. 


\^2t  -  20)dt  =zUt'^  -  20  t 


=  300-176  =  124. 


(h)    r(.^l)5^,=  (^±ll7=64_1^21 


6 

143 


L 


144 


CALCULUS 


EXERCISES 

Evaluate  the  following  definite  integrals 

2 


1. 
3. 

6. 

7. 

9. 

11. 
13. 


2 
dx 

dx 


Vx\x  +  2)dx. 

\    co^-dd. 
Jo         2 

J-il 

Ji     a;  " 

C  2     COS  ^  6?^ 

Jo    l  +  sin2^' 
i      X  sin  2  a;  Ja;. 


^ris.    -\2-. 


vlns.  2.      4, 


/I  TT 

4 

0 


6. 
8. 

10. 


Arts.  ^'    12. 
4 

14. 


J-n  -  X 
r^  jcjix_ 

Jo  4  j^x^' 

\  xe^  dx. 

C^  x^dx 

Jo  a;  +  1 " 

i    arcsin  a;c^a;. 

Jo 

/•log  3 

I    e'^^  dx. 
Jo 


^ns.  log  2. 

Ans.  I  log  2. 

^ns.  2. 

^w.s.  1.568. 


TT 


^4/15.     - 

o 


1. 


^n.s.  4. 


15.  A  body  falls  from  rest  under  gravity.  Find  the  velocity  at 
the  end  of  3  seconds,  and  the  space  described  in  the  third  second. 

16.  A  flywheel,  starting  from  rest,  rotates  under  an  angular 
acceleration  of  tt  radians  per  second  per  second.  Find  the  number 
of  revolutions  made  in  the  fourth  second  of  motion. 

17.  A  point  describes  a  plane  curve,  the  components  of  its  velocity 

at  the  time  t  being 

v^  =  5,  tJy  =  24  -  32 1. 

Find  the  distance  of  the  point  from  its  original  position  at  the  end 

of  2  seconds.  Ans.  18.9  ft. 

99.    Geometric  interpretation  of  a  definite  integraL     It 

was  shown  in  §  81  that  the  indefinite  integral    \f(jc)dx 

represents  the  area  under  the 
curve  y  =  f(x)  between  a  cer- 
tain fixed  ordinate  and  a  vari- 
able ordinate  x  =  x.  In  par- 
ticular, the  change  in  this  area 
as  X  changes  from   a  to    6   is 

-X     the  definite  integral    )  f(^x^dx. 

•/a 

Fig.  57  Hence ; 


0 


^=J  fix)dx 
a 


THE  DEFINITE  INTEGRAL  145 

TTie  definite  integral    I  f^x^dx  may  he  interpreted  as  the 

area  hounded  hy  the  curve  y  =/(a;),  the  x-axis.,  and  the  lines 

X  =  a,  x=  h. 

EXERCISES 

i.  Find  the  area  bounded  by  the  parabola  3/^  =  4  ax,  the  x-axis, 
and  the  lines  x  =  4:a,  x  =  9a.  Ans.  -^/  a^. 

2.  Find  the  area  between  the  curve  i/  =  4:  —  x^  and  the  x-axis. 

3.  Find  the  area  bounded  by  the  curve  ?/  =  log  x,  the  x-axis,  and 
the  line  x  =  2. 

4.  Find  the  area  of  one  arch  of  the  sine  curve. 

100.  Interchanging  limits.  The  effect  of  interchanging 
the  limits  in  a  definite  integral  is  to  change  the  sign  of 
the  integral.     For, 


J 


fCxyx  =  F{b)  -  i^(a), 


jjix)dx  =  F{a-)-F(b); 

i.e.  I  f(x)dx  =  —  I  f(x')dx. 

101.    Change   of    limits    corresponding    to   a   change    of 

variable.     In  the  definite  integral  j  f(x)dx  it  is  of  course 

%y  a 

implied  that  a  and  h  are  the  limiting  values  of  the  variable 
of  integration  x.  If  we  cliange  the  variable  by  a  sub- 
stitution 

x=(j>(z^, 

we  must  either  return  to  the  original  variahle  before  substi- 
tuting the  limits,  or  change  the  limits  to  correspond  with 
the  change  of  variahle.  The  latter  method  is  usually 
preferable. 

The  new  limits  are  found,  of  course,  from  the  equation 
of  substitution 

X  =  (^(2), 


146  CALCULUS 

as  in  the  following 

Example  :    Evaluate    j    x^\  —  x  dx. 

Let 

1— x  =  z^x=l—  z^  dx=  —  dz. 
When 

a:=-l,      2  =  2; 
when 

x=l,  2=0. 

Hence 

j    xVl  —  xdx=  —  j    (1  —  2)2^  6?2 

=  fee*  -  ^4)^^  =  I  ^^  -  I  ^']"=  -  t\V2. 

EXERCISES 

1.   Work  the  above  example,  putting  1  —  x  =  z^. 
Evaluate  the  following  integrals. 

6.    f'     ^^-^      .  .4m.  J^.      7.     T-^^.^ns.  (|V2-2)a. 


8. 


J2  1  _x4'  ®'    Jo  "I 


^   f/x 


a:'^  +  1 

10.  The  area  bounded  by  the  parabola  3/^  =  4  ax,  the  x-axis,  and 
the  latus  rectum  is,  by  §  99, 

A  =  £ydx. 

Evaluate  this  integral  (a)  by  substituting  for  y ;  (b)  by  substituting 
for  dx  and  changing  limits. 

11.  Find  the  area  under  the  curve  y  —  e''  from  x  —  0  to  a:  =  1  by 
the  two  methods  of  Ex.  10. 

12.  Find  the  area  of  half  an  arch  of  the  cosine  curve  by  the  two 
methods  of  Ex.  10. 


THE  DEFINITE  .INTEGRAL  147 

13.  Given  y  =  sin  x,  evaluate   i    xdy  in.  two  ways. 

14.  The  velocity  of  a  point  moving  in  a  straight  line  is 

y  =  4  cos  -  • 
2 

Find  in  two  ways  the  distance  from  the  starting  point  at  the  end  of 

-  seconds, 
o 


CHAPTER    XV 

THE   DEFINITE   INTEGRAL  AS  THE  LIMIT  OF  A  SUM 

102.  Area  under  a  curve.  We  have  seen  in  §  99  that  the 
area  ABQD  bounded  by  the  plane  curve  y  =f(x),  the  a:-axis, 
and  the  lines  x=  a^  x  =  b  is  given  by  the  definite  inte- 
gral  I  f(x)dx.     We  will  now  obtain  a  new  expression  for 

the  same  area. 

In  what  follows,  the  functionj^(a:j  is  assumed  to  be  one- 
valued  and  continuous,  and  to  have  only  a  finite  number^ 
of  maxima  and  minima  in  the  interval  from  x  =  a_to  x==J>; 
in  fact,  we  may  suppose  for  definiteness  that  the  curve 
rises  throughout  the  interval.  The  argument  is  readily 
modified  to  fit  the  case  when  the  curve  steadily  falls, 
or  rises  and  falls  alternately. 

We  may  evidently  get  an   approximate  expression   for 

the  area  A  by  dividing  the 
base  AB  into  n  equal  inter- 
vals Ax,  erecting  the  ordi- 
nates  at  the  points  of  di- 
vision, and  taking  the  sum 
of  the  inscribed  rectangles 
AEFD,  etc.  The  areas  of 
these  rectangles  are  respec- 
tively 


Fig.  58 


where 


f(x^Ax,  f(x^)Ax,'",  f(xJAx, 


X-,  =  a. 


x^  =  a  -\-  Ax, 

Xn  =  a  -\-  2  Ax, 


x^=  a  -{-  (ii  —  1)A2: 
148 


b  —  Ax. 


DEFINITE  INTEGRAL  AS  LIMIT  OF  A  SUM     149 


Hence,  approximately^ 

A  =f{x^}Ax  -^f(x^}Ax  4- 


i=l 


Now  it  is  geometrically  evident  that  this  sum  of  rect- 
angles may  be  made  to  represent  the  area  A  with  an  error 
less  than  any  preassigned  constant  however  small,  by 
taking  n  sufficiently  large.  Hence,  by  the  definition  of 
§  14,  we  have  exactly  —^-^.^ 


(1) 


A=  lim  ^f(Xi)Ax. 


Fig.  59 


A  formal  proof  of  this  state- 
ment may  be  given  as  follows. 
Let  J.J  denote  the  sum  of  the 
inscribed  rectangles,  A^  the  sum 
of  the  circumscribed  rectangles 
(Fig.  59).  Then  for  all  values 
of  n 

A^<  A<  A^. 

Now  the  difference  between  A^  and  A^  is  the  sum  of  the 
shaded  rectangles.  Sliding  all  these  across  into  the  last 
column,  we  see  that 

A^-A^  =  IO.Ax  =  [f(b)  -/(«)]A;r. 

As  n  increases  indefinitely  Ax  approaches  0,  so  that 
A2  —  A^  also  approaches  0.  Since  A  always  lies  between 
A^  and  ^.g,  it  follows  that  A^  and  Ac^  approach  A  as  their 
common  limit.      Hence  we  have  formula  (1).* 

The  rectangles  f(^x^Ax  are  called  elements  of  area.  As 
n  increases  indefinitely,  each  of  the  elements  approaches  0 : 
i.e.  they  are  infinitesimals. 

*  The  n  parts  into  which  AB  is  divided  need  not  be  taken  all  equal; 
the  same  limit  is  obtained  provided  the  width  of  each  rectangle  approaches 
0  as  the  number  of  divisions  is  indefinitely  increased.  Further,  it  is  clear 
that  we  may  take  the  limit  of  the  sum  of  either  the  inscribed  or  the  cir- 
cumscribed rectangles,  or  of  any  set  intermediate  between  these  t^v^o. 


150  CALCULUS 

Questions  like  that  of  the  present  article,  in  which  we 
have  to  deal  with  the  limit  of  a  sum  of  infinitesimal  ele- 
ments, will  arise  many  times  in  this  and  later  chapters. 
As  in  every  case  the  existence  of  the  limit  will  be  evident 
by  geometric  intuition,  we  shall  in  future  omit  formal 
proofs. 

103.  Evaluation  of  the  limit.  Equating  the  values  of  A 
found  in  §§  99  and  102,  we  find 

A  =  lim  y/(a:,)Aa:  =      f(x)dx. 

Thus  the  limit  occurring  in  §  102  can  always  be  evaluated 
by  a  definite  integration.* 

The  fact  that  the  quantity  f(x)dx  appearing  under  the 
integral  sign  represents  the  area  of  a  rectangle  of  altitude 
f(x)  and  base  dx  =  Aa:,  and  thus  suggests  the  sum  from 
which  the   integral  was  derived,  is  the  chief  reason  for 

using  the  notation  j  f(x)dx  (see  §  77).  In  fact,  the  inte- 
gral sign  j  is  historically  a  somewhat  conventionalized  S^ 
meaning  sum. 

104.  The  fundamental  theorem.  In  §  102  we  have  ex- 
pressed the  area  under  a  plane  curve  as  the  limit  of  a  sum 
of  rectangles ;  in  §  99  we  have  found  the  same  area  as  a 
definite  integral.  It  is  clear  that  the  arguments  used 
will  hold  no  matter  ivliat  may  he  the  geometric  or  physical 
meaning  of  the  given  function^  for  any  function  whatever 
may  be  interpreted  as  the  ordinate  of  a  point  on  a  plane 
curve.     We  therefore  have  at  once  the  following 

Fundamental  Theorem  for  Definite  Integrals  : 
Griven  a  function  /(a;),  continuous  in  the  interval  from 
x  =  a  to  X  =  h,  divide  this  interval  into  n  equal  parts  A2:, 

*  More  precisely,  the  limit  can  always  be  expressed  as  a  definite  inte- 
gral; the  actual  evaluation  of  the  integral  is  often  impossible  (see  §  78). 


DEFINITE  INTEGRAL  AS  LIMIT  OF  A  S\jM     153 

n 

and  form  the  sum  ^f(^Xi)Ax^   where  x^  =  a,  x^  =  a  - 
. ..,  Xn='  a  +(n—  l)Aa;.      The7i 


Hence^  if  in  any  problem  an  arbitrarily  close  approxima- 
tion *  to  the  required  quantity  can  be  found  by  adding  up 
terms  of  the  typef(x')Ax  from  x  =  a  to  x  =  b^  that  quantity 

is  given  exactly  by  the  definite  integral    I    f{x)dx. 

We  have  now  presented  the  ^efiiiite  integral  in  two 
distinct  aspects  :  first,  as  the  change  in  the  value  of  tHe 
indefinite  integral  between  two  values  of  the  variable ; 
second,  as  the  limit  of  a  sum  of  infinitesimal  elements.  The 
great  advantage  of  this  latter  point  of  view  will  become 
apparent  as  we  proceed. 

It  may  be  remarked  that  if 
the  function  f(x^  has  a  finite 
number  of  finite  discontinuities 
in  the  interval  from  a  to  6,  as 
in  Fig.  60,  it  can  still  be  inte- 
grated. For  it  will  be  con- 
tinuous in  a  number  of  sub-  ^^^'  ^ 
intervals  such  as  AO^  QD^  DB^  to  each  of  which  the 
fundamental  theorem  can  be  applied  and  the  results  added. 

105.  Plane  areas  in  cartesian  coordinates.  Not  only  the 
area  considered  in  §  102,  but  any  plane  area  bounded  by 
curves  whose  equations  are  given  in  cartesian  coordinates, 
can  be  found  as  follows.  Imagine  inscribed  in  the  area  a 
set  of  n  elementary  rectangles  of  altitude  h^  and  width  AZ, 

n 

in  such  a  way  that  the  sum  V  ^^A^  may  be  made  to  repre- 


<=i 


*  That  is,  an  approximation  in  which  the  error  may  be  made  less  than 
any  preassigned  constant. 


150 


CALCULUS 


Quer^e  area  to  any  desired  degree  of  approximation  by 
hav"  rising  n.     Then  at  once,  by  the  fundamental  theorem, 

A=  lim   V7z,A/=  ChdU 

the  limits  being  chosen  in  such  a  way  as  to  extend  the 
integration  over  the  whole  area.  Of  course  in  any  par- 
ticular problem  li  and  dl  must  be  ex- 
pressed in  terms  of  the  coordinates. 

In  every  problem  the  student  should 
make  a  sketch  of  the  area  to  be  found, 
draw  an  element  in  a  general  position, 
and  obtain  the  area  of  the  element  directly 
from  the  figure,  as  in  the  following 

Examples:  (a)  Find  the  area  in  the 
first  quadrant  bounded  by  the  parabola 
?/2  =  4:ax,  the  a^-axis,  and  the  line  x—  a. 

An  arbitrarily  close  approximation  to 
this  area  can  be  found  by  forming  a  sum 
of  rectangles  of  altitude  y,  base  dx^  and  area  y  dx^  as 
shown  in  Fig.   61.      Hence,  we  have  exactly 

ra  /»«  2      2         i' 

A=  \    y  dx  =  2  \     \:ax  dx  =  ~  •  -  (ax^  ^ 
Jo  "^  Jo     [  ^a     3^     ^ 

(K)  Find  the  area  in  the  first  quadrant 
between  the  parabolas 

(1)  y^  =  -^  ^^, 

(2)  ?/2  =  8a2:- 4^2. 

Let  us  take  as  the  element  a  rectangle 
parallel  *  to  OX.  The  area  of  the  rect- 
angle is  evidently  (x^  —  x^dy,  where  x^ 
and  x^  are  the  abscissas  of  the  points  on 
the  curves  (1)  and  (2)  respectively.     The  fig.  62 


Fig.  61 


=  -a^ 


*That  is,  with  its  finite  side  parallel  to  OX. 


DEFINITE  INTEGRAL  AS  LIMIT  OF  A  SUM     153 
curves  are  found  to  intersect  at  (a,  2  a).      Hence 

A  =r(r,^  -  X,)  dy  =r(f  +  5  -Pjdy 
*^o  ^0     V8  a      2      4  aj 


'£ 


\:^      SaJ  ^     12       24a 


2 

=  -  a^. 
3 


EXERCISES 

1.  Find  the  area  bounded  by  the  curve  y  =  x^  and  (a)  the  lines 
y  =  0,  X  =  2,  (h)  the  lines  x  =  0,  y  =  1. 

2.  Solve  example  (a),  §  105,  taking  the  element  parallel  to  OX. 
Evaluate  the  integral  in  two  ways. 

3.  Find  the  area  bounded  by  the  curve  ay'^  =  x^  and  the  line 
X  =  '^a.  Ans.   i|^  a"^. 

4.  Find  the  area  of  a  circle  (see  Ex.  52,  p.  136). 

5.  Find  the  area  of  an  ellipse,  using  the  cartesian  equation ;  check 
by  using  the  equations  x  =  a  cos  (f>,  y  —  b  sin  cf)  (see  Ex.  51,  p.  136). 

A  ns.    irab. 

6.  Find  the  area  of  half  an  arch  of  the  curve  y  =  I  cos  2  x. 

7.  Find  the  area  between  the  parabolas  y^  =  4:ax  and  x^  =  4  ay.  ■ 

A71S.  1/  a-. 

8.  Show  that  the  area  bounded  by  a  parabola  and  any  chord 
perpendicular  to  the  axis  is  two  thirds  of  the  circumscribing 
rectangle. 

9.  Find  in  two  ways  the  area  bounded  by  the  parabola  y  =  x'^,  the 
?/-axis,  and  the  lines  y  =  1,  y  =  4. 

10.  Find  the  area  bounded  by  the  curve  y  —  log  x,  the  axes,  and  the 
line  ?/  +  1  =  0. 

11.  Find  the  area  bounded  by  the  curve  y  =  log  x,  the  x-axis,  and 
the  line  x  =2.  Ans.  0.386. 

a  I   -       — ^\ 

12.  Find    the    area    under    the    catenary    ?/=-(ga+e    aj,    from 

X  —  —  a  \iO  X  =  a.  An^.  aP-\  e 


e 

13.  Find  the  area  between  the  curve  x^  =  4  a^  —  ay  and  the  x-axis, 

taking  the  element  («)  parallel  to  0Y\  {h)  parallel  to  OX. 

Ill 

14.  Find  the  area  bounded  by  the  parabola  X'^  -\-  y'^  =  a'^  and  the 

coordinate  axes.  Ans.  ^a^. 


154  CALCULUS 

15.  Find  the  area  of  a  circular  segment  of  height  h.     Check  by- 
putting  A  =  2  r. 

16.  Find   the  area   of    one    arch    of   the  cycloid  x  =  a(6  —  sin  6), 
y  =  a(l  —  cos  6).  Ans.  3  na^. 

17.  Find  the  area  bounded  by  the  curves  y  =  x,  y  =  2  x,  y  =  x^. 

Ans.  |. 

18.  Find  in  two  ways  the  area  in  the  first  quadrant  bounded  by  the 
curves  y  =  x^,  x'^  =  2  —  y,  y  =  0. 

19.  Find  in  two  ways  the  area  bounded  by  the  curve  y=(l  —  x^y 
and  the  a;-axis. 

20.  Find  the  area  bounded  by  the  curve  y  =(x  —  Sy^(x  —  2)  and 
the  a'-axis. 

21.  Find  the  area  bounded  by  the  curve  y  =  —^ — -,  its  asymptote, 
and  the  maximum  ordinate.  Ans.  0.698. 

22.  Trace  the  curve  y^(x'^  +  a"^)  =  aP'x'^,  and  find  the  area  bounded 
by  the  curve  and  the  line  x  =  a.  Ans.  0.83  a^. 

23.  Trace  the  curve  ay-  =  ax^  +  x^,  and  find  the  area  of  the  loop. 

24.  Find  the  area  bounded  by  the  curve  y^  =  x^(l  +  x)  and  the 
ar-axis.     Why  is  the  answer  negative  ? 


25.  Trace  the  curve  y  = ;- ,  and  find  the  area  under  the  curve 

from  X  =  -  2  to  X  :ir  0.  ^  +  ^^  ^^,^  j  37_ 

26.  Find  in  two  ways  the  area  bounded  by  the  coordinate  axes  and 
the  curve  y^  =  1  —  2  x  —  xy. 

27.  Find  the  area  of  the  circle  x  =  a  sin  2  6,  y  =  a  cos  2  6. 

28.  Find  the  area  bounded  by  the  curve  y  =  — ^—,  the  a:-axis,  and 
the  maximum  ordinate.  Ans.  ^. 

29.  Find  the  area  in  the  first  quadrant  under  the  curve  y^  = , 

X  —  I 

between  the  minimum  ordinate  and  the  ordinate  at  a;  =  3.        Ans.  2.05. 

30.  Trace  the  curve  y^  =  x'^(x  +  4:),  and  find  the  area  inclosed  by  it. 

212 

Ans.  — — . 
105 

106.    Plane  areas  in  polar  coordinates.     Given  the  equa- 
tion 

of  a  plane  curve  in  polar  coordinates,  let  us  try  to  find  the 
area  bounded  by  the  curve  and  the  radii  vectores  corre- 


DEFINITE  INTEGRAL  AS  LIMIT  OF  A  SUM     155 

spending  to  6  =  a,  6  =  /3.  We  can  obtain  an  arbitrarily 
close  approximation  to  the  area  by  inscribing  in  it  n  circu- 
lar sectors  of  radius  r^  and  angle  A^, 
hence  of  area*  |  r,^A^,  and  forming 

n 

the  sum  V  J  r,^A6.     Hence,  by  the 
fundamental  theorem, 


i=l 


By    means    of   a   theorem    to    be  ^^^-  ^^ 

proved  in  §  109,  it  can  be  shown  that  this  result  may 
also  be  obtained  by  choosing  as  the  element  a  triangle 
of  altitude  r^,  base  rjA^,  and  area  J  r'jAB. 


EXERCISES 

1.  Find  the  area  swept  out  by  the  radius  vector  of  the  spiral  of 
Archimedes  r=a6,  in  the  interval  from  ^  ==  0  to  ^  =  2  tt. 

2.  Solve  Ex.  1  for  the  logarithmic  spiral  log  r  =  a6. 

3.  Find  by  integration  the  area  of  the  triangle  bounded  by  the  lines 

r  =  a  sec  0,  B  =  (),  6  =  —  • 

4 

4.  Find  the  area  inside  the  lemniscate  r^  =  a^  cos  2  6.  Ans.  a^. 

5.  Find  the  area  of  the  curve  r^  =  a^  cos  $. 

6.  Find  the  entire  area  of  the  cardioid  r  =  a(l+  cos  0) .     Ans.  f  ira^. 

n 

7.  Find  the  area  between  the  parabola  r  =  a  sec'-^-  and  its  latus 

rectum.  "  Ans.^a^. 

8.  Find  the  area  of  the  curve  r  =  a  sin 2  6.  Ans.  ^wa^. 

9.  Show   that  the  area  of  one  loop  of  the   curve  r  =  a  cos  nO  is 

2 

,  hence  the  total  area  inside  the  curve  is  one  fourth  or  one  half  the 

4  71 

area  of  the  circumscribed  circle,  according  as  7i  is  odd  or  even. 

*  By  elementary  geometry,  the  area  of  a  circular  sector  of  radius  r 
and  angle  a  is 

A  =  l  r^a. 


156 


CALCULUS 


107.    Volumes  of  revolution.     The  volume  of  a  solid  of 
revolution  may  be  found  very  readily  by  the  fundamental 

theorem. 

Suppose  the  volume 
is  generated  by  revolv- 
ing the  area  ABCD 
about  the  line  AB.  If 
we  inscribe  in  the  re- 
volving area  a  set  of 
n  rectangles  of  alti- 
tude r»  and-  base  AA, 
each  rectangle  will  gen- 
erate in  its  rotation  a 
circular  disk^  or  cyl- 
inder, of  radius  r^,  alti- 
tude Ah,  and  volume 
irr^Ah.  Further,  as  n 
increases  the  sum  of 
these  cylindrical  vol- 
umes approaches  as  its 
limit  the   required  vol- 


FiG.  64 


b 


ume.      Hence,  by  the  fundamental  theorem, 


i=l 


the  limits  being  chosen  so  as  to  include  the  whole  volume. 
Of  course  in  any  problem  both  rand  dh  must  be  expressed 
in  terms  of  the  coordinates. 

When  the  axis  of  revolution  does  not  form  part  of 
the  boundary  of  the  revolving  area,  we  may  choose  as 
elements  a  set  of  circular  rings,  as  in  example  (6) 
below. 

Examples :  (a)  The  area  in  example  (a),  §  105,  revolves 
about  OX.     Find  the  volume  generated. 

Dividing  the  area  into  elements  as  in  Fig.  61,  we  see 


DEFINITE  INTEGRAL  AS  LIMIT  OF  A  SUM     157 


that  each  rectangle  generates  a  cylindrical  volume-element 
of  radius  ?/,  altitude  dx^  and  volume  iry'^dx.     Hence 


V 


="X' 


y^ dx  —  \iTa\    xdx=  27ra^. 

(6)  The  above  area  rotates  about  OY. 
Find  the  volume  generated. 

If  we  divide  the  area  into  elements  as 
in  the  figure,  each  element  generates  a 
circular  ring  of  outer  radius  a,  inner 
radius  x^  thickness  dy^  and  volume 
ir^a^  —  ^^dy.  Further,  the  limit  of  the 
sum  of  these  volumes  is  the  required 
volume.     Hence 

F=  'rr£\a'^  -  x^)dy  =  ttJ'Y^^  - 


Fig.  65 


y' 


16  a^ 


dy 


r_ 

80  a^ 


=  -Tra*- 


This  result  could  have  been  obtained  equally  well  by 
finding  the  volume  generated  by  rotating  the  area  OBQ 
about  Oy,  and  subtracting  this  from  the  volume  of  the 
cjdinder  formed  by  revolving  the  rectangle  OABQ. 
But  in  case  it  is  possible  to  simplify  the  integral  before 
integrating,  as  often  happens,  the  first  method  is  to  be 
preferred. 

108.  Volumes  of  revolution :  second  method.  The  fol- 
lowing method  for  finding  volumes  of  solids  of  revolution 
is  often  preferable   to  that  of  §  107. 

Let  us  take  as  an  element  of  the  area  ABO  (Fig.  (^6^  a 
rectangle  of  length  hi  parallel  to  the  axis  of  revolution  AB^ 
and  of  width  Ar.  This  rectangle  generates  by  its  rotation 
a  cylindrical  shell  of  inner  radius  r^,  altitude  A^,  and  thick- 
ness Ar.     The  volume  of  the  shell  is  evidently 

7r(ri  -t-  Ar')Vii  —  irrfhi  =  2  TrrihiAr  +  Tr/^^Ar^, 
and  the  limit  of  the  sum  of  these  elementary  shells  is  evi- 


158 


CALCULUS 


dently  the  required  volume.     Now  it  will  be  shown  in  the 
next  article  that,  in  passing  to  the  limit,  we  may  neglect 

the  infinitesimal  of  higher 
order*  7rhiAr\  Hence,  by 
the  fundamental  theorem, 


Fig.  66 


II 

V=  Urn  ^  2  TTTihiAr 

=  2Tr  Crh  dr, 

the  integration  being  ex- 
tended through  the  whole 
region. 

This  result  is    easily  re- 
membered   from    the    fact 
that  the  integrand  is    the 
differential    of    volume    of 
a  right  circular  cylinder,  the  altitude  being  constant: 
F=7rr2A,    dV=2  7rrhdr. 
Example :     Solve  example  (6),   §  107,  by  the  present 

method. 

Divide  the  rotating  area  into  rectangles  parallel  to  OF, 
as  in  Fig.  61.  Each  rectangle  generates  a  cylindrical 
shell  of  radius  x,  altitude  g,  and  thickness  dx.     Hence 

V=  2  TT  j    xg  dx  =  4:  ira^  I    x^  dx  =  f  ira^. 

109.  A  theorem  on  infinitesimals.  It  often  happens,  as 
in  the  preceding  article,  that  in  applying  the  fundamental 

n 

*  This  is  easily  shown  directly.    The  quantity  neglected  is  ^  vhi^r^. 

n  i=l 

This  may  be  written  irAr  ^  hi^r.      When   Ar  approaches  0,  the  sum 

n  1=1 

^  hiAr  approaches  a  finite  limit,  viz.  the  generating  area,  so  that  the 

1=1 

whole  quantity  approaches  0. 


DEFINITE  INTEGRAL  AS  LIMIT  OF  A  SUM     159 

theorem  we  have  to  replace  the  element  as  originally 
chosen  by  another  element  differing  from  the  first  one  by 
an  infinitesimal  of  higher  order.  That  this  is  allowable 
appears  from  the  following 

Theorem;  The  limit  of  a  sum  of  positive  infinitesimals 
is  unchanged  when  each  infinitesimal  is  replaced  hy  another 
that  differs  from  it  hy  an  infinitesimal  of  higher  order. 

It  follows  that,  in  taking  the  limit  of  such  a  sum,  all  in- 
finitesimals of  higher  order  may  he  neglected^  as  was  done  in 
§108. 

In  this  connection  it  should  perhaps  be  mentioned  ex- 
plicitly that  two  infinitesimals  differ  from  each  other  by 
an  infinitesimal  of  higher  order  whenever  the  limit  of  their 
ratio  is  1,  and  conversely. 

To  prove  the  theorem,  let  Wj,  2^2'   ••'  ^n  be  a  set  of  posi- 

tive  infinitesimals  such  that  lim  ^  ^»  exists;   and  let  Vj, 

^2,  •••,  v„  be  another  set  such  that  Vi  differs  from  u^  by  an 
infinitesimal  of  higher  order : 

^t  =  "^i  +  «^t?/t, 

where  w^  is  infinitesimal.     Then 

n  n  n 

lim  2  Vi=  lim  ^Ui-^  lim  ^  ?/\Wj. 

Denoting  by  w  the  absolute  value  of  the  largest  of  the  t^'s, 
we  have 

w  n  n 

-W^Ui<^  Willi  ^W^Ui. 
t=l  i—\  t=l 

Since  the  first  and  third  of  these  quantities  both  approach 
0,  the  second  must  do  likewise.     Hence 

n  n 

lim  ^  Vi  —  lim  ^  w^, 
and  the  theorem  is  proved. 


160  CALCULUS 

EXERCISES 

1.  Find  the  volume  of  a  sphere. 

2.  Find  the  volume  of  a  right  circular  cone. 

3.  The  hyperbola  x^  —  y'^  =  a^  revolves  about  its  transverse  axis. 
Find  the  volume  of  a  segment  of  height  a  of  the  hyperboloid  gen- 
erated. Ans.    |7ra3. 

4.  Find  the  volume  generated  by  revolving  the  four-cusped  hypo- 
cycloid  a;3  +  3/3  _  ^3  about  OX.  Ans.  j%%  ira^ 

5.  Find  the  volume  generated  by  revolving  the  area  under  the 
curve  ?/  =  e^  from  x  =  0  to  x  =  1  (a)  about  OX;  {b)  about  OY;  (c) 
about  the  line  x  =  1.  Ans.     (c)  2  7r(e  —  2). 

6.  The  area  under  one  arch  of  the  sine-curve  revolves  (a)  about 
OX;  (b)  about  OF.     Find  the  volume  generated. 

Ans.    (a)   ^i   (b)  2  7r2. 

7.  Find  the  volume   obtained  by  revolving  about  OX  the  area 

X  X 

under  the  catenary  ?/  =  -(ga  +  e    «),  from  x  =  —  a  to  x  =  a. 


Ans.    — 7- 
4 


V^'-i) 


8.  The  area  OBC  of  Fig.  65  revolves  (a)  about  the  line  CB ; 
(b)  about  AB.  Find  the  volume  generated;  check  by  solving  in 
two  ways. 

9.  Find  the  volume  of  an  oblate  spheroid,  using  («)  the  ordinary 
equation  of  the  ellipse;  (6)  the  parametric  equations  a;  =  acos<^, 
y  =  bsm^.     Solve  each  part  in  two  ways.  A71S.   f  Tra^ft. 

10.  The  area  in  example  (b),  §  105,  revolves  about  OX.  Find  the 
volume  generated. 

11.  Find  the  volume  of  a  spherical  segment  of  height  h.  Check 
by  putting  h  =  2  r. 

12.  Trace  the  curve  a'^y'^  =  x^(2  n  —  x),  and  find  the  volume  gen- 
erated by  revolving  the  curve  about  the  a:-axis. 

13.  Find  the  volume  generated  by  revolving  (a)  about  OX,  (b) 
about  OY,  the  area  between  the  curves  2  y  =^  x^,  y  =  x'^.  Check  the 
results  by  solving  in  two  ways. 

14.  Trace  the  curve  (x  —  4  a)y'^  =  ax(x  —  3  a),  and  find  the 
closed  volume  generated  by  revolving  it  about  the  a:-axis. 

Ans.    6.12  a3. 


DEFINITE  INTEGRAL  AS  LIMIT  OF  A  SUM     161 


15.    The    curve    y^  =  x(x  —  l)(x 
Find  the  closed  volume  oeuerated. 


2)    rotates   about    the   a;-axis. 


A  ns.    —  • 
4 


16.  Find  the  volume  generated  by  revolving  one  arch  of  the  cy- 
cloid X  =  a(0  —  sin  0),  ij  =  a(l  —  cos  0)  about  the  ar-axis.    Ans.  5  ir^a^. 

17.  Trace  the  curve  y(^x^  +  y^)  =  a(^"^  —  y'^),  and  find  the  volume 
generated  by  revolving  the  loop  about  OY.  Ans.   0.053  ira^. 

18.  Find  the  volume  of  a  torus.     Solve  in  two  ways.     Ans.  2  Tr^a%. 

19.  The  area  bounded  by  the  curve  y  =  (1  —  x-)^  and  the  a:-axis 
revolves  about  the  ?/-axis.  Find  the  volume  generated,  (a)  by  the 
method  of  §  107 ;  (h)  by  the  method  of  §  108.  In  (a)  evaluate  the 
integral  in  two  ways,  first  by  substituting  for  x^,  next  by  substituting 
for  dy. 

20.  Trace  the  curve  a^y'^  =  a^x'*^  —  x^,  and  find  the  volume  gener- 
ated by  revolving  one  loop  about  OY.  Ans.   ^^  ira^. 

21.  A  round  hole  of  radius  a  is  bored  through  the  center  of  a 
sphere  of  radius  2  a.     Find  the  volume  cut  out. 

22.  Find   the    closed   volume   generated   by  revolving   the   curve 

X^Cx^  Q,   ^ 

y~  = — ^, . — T-    about  the  x-axis.     Trace  the  curve.     Ans.  0.072  ttci^. 

23.  Find  the  volume  inside  the  cylinder  x-  +  ?/-  =  2  a^  and  outside 
the  hyperboloid  x^  +  y'^  —  z^  —  a'^.  A7is.   |  ira^. 

24.  Find  the  closed  volume  generated  by  revolving  the  curve 
y-  =  x-^(x  +  4),  (a)  about  the  x-axis,  (b)  about  the  ^/-axis. 

25.  Find  in  two  ways  the 
volume  generated  by  revolv- 
ing about  the  ?/-axis  the   area 

bounded  by  the  curve  y  = 

X 

and  the  coordinate  axes. 

110.    Other   volumes. 

The  volume  of  any  solid 
can  be  expressed  as  a  defi- 
nite integral,  provided  we 
know  the  area  of  every 
plane  section  parallel  to 
some  fixed  plane.  Let  us 
divide  the  volume  into 
slices  of  thickness  Ah  by 

M 


Fig.  67 


162 


CALCULUS 


means  of  n  planes  parallel  to  this  fixed  plane.  If  on  each 
of  these  plane  sections  we  erect  a  cylinder  of  altitude  Ah 
and  base  A^  where  A  is  the  area  of  the  section,  the  sum 
of  the  n  cylindrical  volumes  thus  formed  will  be  approxi- 
mately the  required  volume,  and  the  limit  of  this  sum 
will  be  exactly  the  volume.     Hence 


V=    limXAi^h  =  fAdh, 

n-^QC  t=l  *^ 


with  properly  chosen  limits. 

Example :  A  woodsman  chops 
halfway  through  a  tree  4  ft.  in 
diameter,  one  face  of  the  cut 
being  horizontal,  the  other  in- 
clined at  45°.  Find  the  volume 
of  wood  cut  out. 

The  figure  shows  one  half  of 
the  required  volume.  If  we  slice 
up  the  volume  by  planes  parallel 
to  the  ^2-plane,  the  element  of 
volume  is  a  triangular  plate  of 
width  1/,  altitude  2,  and  thickness 
dx.     Hence 

V=  2  Pj  ^z  dx. 
But 


Fig.  68 


so  that 


2  =  y,  and  2/  =  V4  — 


X' 


V=  r(4:-x^}dx=5^cu. 

Jo 


ft. 


EXERCISES 

1.  Solve  the  above  example  in  a  different  way. 

2.  Find  the  volume  of  a  tetrahedron  with  three  mutually  perpen- 
dicular faces.  Ans.   ^abc. 

3.  Find  the  volume  sliced  off  from  a  right  circular  cylinder  by  a 
plane  through  a  diameter  of  one  base  and  tangent  to  the  other  base. 

Ans.    I  rt^A. 


DEFINITE  INTEGRAL  AS  LIMIT  OF  A  SUM     163 

4.  Find  the  volume  of  a  right  pyramid  with  a  square  base. 

5.  Find  the  vohime  of  an  ellipsoid,  using  the  answer  to  Ex.  5, 
p.  153.  Alls,   ^irabc. 

6.  Find  the  volume  of  an  elliptic  cone.  Ans.   ^  Trabh. 

7.  Find  the  volume  of  a  spherical  wedge.  Ans.    f  aa^. 

8.  Find  the  volume  of  a  wedge  cut  from  a  right  circular  cone  by 
two  planes  through  the  axis.  Ans.   i  aa%. 

9.  Obtain  a  formula  for  the  volume  of   a  wedge  cut  from  any 
solid  of  revolution  by  two  planes  through  the  axis. 

10.  A  carpenter  chisels  a  square  hole  of  side  2  in.  through  a  round 
post  of  radius  2  in.,  the  axis  of  the  hole  intersecting  that  of  the  post 
at  right  angles.     Find  the  volume  of  wood  cut  out.     Ans.   15.3  cu.  in. 

11.  Find  the  volume  cut  from  the  cylinder  x'^  +  t/'^  =  a^  by  the 
planes  z  =  ?nx,  z  =  nx.     Solve  in  two  ways. 

12.  A  right  circular  conoid  is  generated  by  a  straight  line  which 
moves  always  parallel  to  the  a:y-plane  and  passes  through  the  line  y  =  h 
in  the  ?/2-plane  and  the  circle  x'^  -{-  z^  =  a^  in  the  xs-plane.  Find  the 
volume  of  the  conoid.  Ans.    |  ira^. 

13.  Find  the  volume  in  the  first  octant  bounded  by  the  hyperbolic 
paraboloid  generated  by  a  straight  line  moving  always  parallel  to  the 
z?/-plane  and  passing  through  the  lines  y  +  z  =  a  in  the  yz-plane  and 
X  =  b  in.  the  a^s-plane.  Ans.    ^  a%. 

14.  A  banister  cap  is  bounded  by  two  equal  cylinders  of  revolution 
8  in.  in  diameter,  whose  axes  intersect  at  right  angles  in  the  plane  of 
the  base  of  the  cap.     Find  the  volume  of  the  cap  in  two  ways. 

15.  Find  the  volume  of  a  right  pyramid  whose  base  is  a  regular 
hexagon. 

16.  Find  the  volume  in  the  first  octant  under  the  plane  z  =  x  and 
inside  a  cylinder  standing  on  the  parabola  ?/  =  4  —  a;^  as  a  base.  Solve 
in  two  ways. 

17.  Solve  Ex.  12  if  the  line  ?/  =  A  is  replaced  by  the  line  y  -\-  z  =  h 
(/i>a).  Ans.  ^TTO^h. 

18.  Solve  Ex.  13  if  the  line  x  =  b  is  replaced  by  the  line  x  =  z. 

19.  Find  the  volume  in  the  first  octant  bounded  by  the  planes 
X  ■\-  z  =  a,  X +  2^  +  22  =  2 a. 

111.  Line  integrals.  The  ordinary  definite  integral 
depends  on  all  the  values  of  a  given  function /(a:)  along 
a  straight  line  segment  —  the  segment  of  the  a;-axis  from 


164 


CALCULUS 


Fig.  69 


x=  a  to  X  =  h.  It  happens  frequently  that  we  have  to 
compute  a  quantity  that  depends  in  a  similar  way  on  the 
values  of  a  function  F  along  a  curvilinear  arc  0.  The 
function  F  is  in  general  dependent  on  both  coordinates  of 
the  point  on  the  curve  : 

F=  Fix,  y). 

But  since  y  is  given  as  a  function  of  x  by  the  equation  of 
the  curve,  the  function  F(x^  y)  reduces  at  once  to  a  func- 
tion of  one  variable. 

Given  a  function  F(x,  y)  de- 
fined at  all  points  of  a  plane 
curve  (7,  let  us  inscribe  in  C  a 
broken  line  of  n  segments  As/ 
having  equal  projections  ^x  on 
the  a:-axis,  multiply  each  seg- 
ment by  the  value  of  F(x^  ^/)  at 
the  corresponding  point  of  division*  P^  (Fig.  69),  and  form 

n 

the  sum  of  these  products,  2j  -^fe^  Vi)  ^^l '     The  limit  of 

this  sum,  as  the  number  of  divisions  becomes  infinite,  is 
called  the  line  integral  of  F(^x^  y)  along  the  arc  (7,  and  is 

denoted  by  the  symbol   j  F{x^  y)  ds  : 

lis.  %  ^(^i'  y^^'l  =£^(^^  y)  <^'- 

112.    Geometric  interpretation  of  the  line  integral.     The 

existence  of  the  limit  last  written  may  be  made  evident 
geometrically.     Let  us  interpret  the  function  F(^x^  y)  as 
the  2-coordinate  of  a  point  on  a  surface  in  space  : 
(1)  z  =  F(ix,y). 

*  It  is  of  course  merely  for  convenience  that  the  broken-line  segments 
are  drawn  with  equal  projections  on  OX.  Tlie  division  may  be  made  in 
any  manner  provided  in  the  limit  every  segment  approaches  0.  Further, 
Asj'  may  be  multiplied  by  the  value  of  i^(x,  y)  at  either  end-point  of  ASj' 
or  at  any  point  on  the  subtended  arc. 


DEFINITE  INTEGRAL  AS  LIMIT  OF  A  SUM     165 


On  the  curve  C  as  directrix  erect  a  cylindrical  sur- 
face with  generators  perpendicular  to  the  rr^z-plane. 
Each  of  the  quan- 
tities F(Xi,yi)  As/ 
represents  the  area 
of  a  rectangle  in- 
scribed in  this  cyl- 
inder,  having  a 
base  As/  and  an  al- 
titude F(x^^  1/i),  and 
the  sum  of  these 
rectangles  evi- 
dently approaches 
as  its  limit  that 
part  of  the  cylin- 
drical surface  ly- 
ing between  the  a;^-plane  and  the  surface  (1). 

113.  Fundamental  theorem  for  line  integrals.  The 
theorem  of  §  104  takes  the  following  form  for  line  in- 
tegrals : 

Given  a  function  FQx^  ?/)  defined  at  all  points  of  an  arc 
C,  iyiscrihe  in  the  arc  a  broken  line  of  n  segments  As/  having 

n 

equal  projections  on  OX,  and  form  the  sum  ^  F{Xi,  t/i)  ASi, 
where  (xi,  y^  is  the  i-th  point  of  division  on  C.      Then 
Urn  V  Fix^.  y^)As^  =  (f(x,  y)ds. 


Fig.  70 


n-^cc 


t=l 


Hence,  if  in  any  problem  an  arbitrarily  close  approxima- 
tion to  the  required  quantity  can  be  found  by  addi?ig  up 
terms  of  the  type  F(^x,  y')As',  the  quantity  is  given  exactly 

114.  Evaluation  of  line  integrals.  To  evaluate  a  line 
integral,  we  have  in  general  to  express  both  F(^x,  ?/)  and 
ds  in  terms  of  x,  y,  or  some  other  suitable  variable,  and 


166  CALCULUS 

then  integrate  between  limits  in  such  a  way  as  to  extend 
the  integration  over  the  given  arc. 

Thus,  if  X  is  chosen  as  the  variable  of  integration,  we 
replace  ds  by  its  value  (§  52) 


^^=Vi+(tT^-' 


\dx) 
and  obtain 

JFix,  y-)d»=£Fix,  y)^ll  +(^)V 

where  a  and  b  are  the  abscissas  of  the  end  points  of  C,* 
and  where  y  must  be  replaced  by  its  value  in  terms  of  x 
from  the  equation  of  the  curve,  f 

-  *  It  is  assumed  that  no  parallel  to  the  y-Rxis  can  meet  C  in  more  than 
one  point.  If  this  condition  is  not  satisfied,  C  must  consist  of  several 
portions  for  each  of  which  the  condition  holds,  and  each  portion  may  be 
considered  separately. 

t  This  transformation  of  the  line  integral  into  an   ordinary  integral 
may  be  justified  by  the  theorem  of  §  109.     We  have  evidently 

ASi'  =  Va^'  +  A^'  =  \  1  +  ^Ax. 

^  Ax 

Now  the  limit  of  the  sum   ^  F  (Xi,  Pi)^!  +  ^^  Ax  cannot  be  ex- 

!=i  Ax^ 

pressed  directly  as  a  definite  integral  by  the  theorem  of  §  104,  since  the 
summand  depends  not  only  on  Xi  but  on  Ax  as  well.     But  the  infinitesi- 


mals F(Xi,  ?/i)\/l  -\--=2^^  ^^^  F{Xi,  yi)y/l  +  y''^ Ax  differ  from  each 


Ax 


other  by  an  infinitesimal  of  higher  order,  since  the  limit  of  their  ratio  is 
evidently  1,  and  hence  the  latter  may  be  substituted  for  the  former,  by 
§  109.     Therefore 

\  F{x,  y)ds  =   lim    V  F(x,,  ?/,)\/l  +  ^  Ax 
JC  n->oo  ^  ^  Ax^  ■ 

n 

=   lim    V  F(Xi,  yi)Vl+yi'2Ax 
=  Cf(x,  7j)y/l  +  y'-^dx, 

Ja 

by  the  fundamental  theorem  of  §  104. 


DEFINITE  INTEGRAL  AS  LIMIT  OF  A  SUM     167 
To  integrate  with  respect  to  ^,  we  put 

and  express  the  entire  integrand  in  terms  of  y. 

If  X  and  y  are  given  in  terms  of  a  parameter  t^  we  use 


In  the  discussion  of  line  integrals,  we  have  spoken  in 
terms  of  cartesian  coordinates,  but  the  argument  is  evi- 
dently independent  of  the  particular  coordinate  system 
used. 

Some  simple  types  of  line  integrals  are  considered  in 
the  next  three  sections  ;  other  examples  will  be  met  with 
later. 

115.  Length  of  a  curvilinear  arc.  To  find  the  length  of 
an  arc  of  a  plane  curve  (7,  we  proceed  as  follows  :  Inscribe 
in   Q  a  broken  line  of  n  segments  As/  as  in  §  111,  and 

n 

form  the  sum  V  As/.     This  sum  is  of  course  the  length  of 

the  broken  line,  and  its  limit  is  the  length  s  of  the  arc.  It 
is  evidently  the  line  integral  I  c?s,  the  given  function  in 
this  case  being  ^{x^  y')-—!: 

s=  lim   VAsJ=fds. 


The  process  of  finding  the  length  of  a  curve  is  some- 
times called  rectification  of  the  curve. 

Example :  Find  the  circumference  of  the  circle 

x^  -\-  y^=  a^. 
Here 

dy  _  _x 
dx         y^ 


168 
so  that 


CALCULUS 


i      C       dx 
=  4a  I     — r^:= 


X 


=  4  a  arcsin  - 

Va2  -  x^  «. 


=  2  7ra. 


EXERCISES 

1.  Find  the  circumference  of  the  circle  x  =  a  cos  9,  y  =  a  sin  ^. 

2.  Rectify  the  semicubical  parabola  ay^  =  x^  from  a:.=  0  to  a:  =  5  a. 

Ans.  -^T^j^-a. 

3.  Trace   the   curve   9  y"^  ■=  4:(1  +  x-y,   and   find   its   length    from 


i:  =  0  to  a;  =  2. 

4.    Rectify   the   catenary   3/ =  -  [e«  +  e~o  J 


ylm.  -232. 


e  <» 

2         2 


from  a:  =  0  to  a:  =  a: 
2\ 

2.  L  ^ 

5.  Find  the  length  of  the  four-cusped  hypocycloid  x^  -^  y^  =  a^ 

Ans.  6  a. 

6.  Rectify  the  curve  x  =  t%  y  =  fi  from  ^  =  0  to  ^  =  VS. 

5 

7.  Find  the  length  of  the  curve  y  =  ^x^  between  the  origin  and 

the  point  a;  =  4.  Ans.     ^^^(l+ev^). 

8.  Find  the  length  of  one  arch  of  the  cycloid.  Ans.  8  a. 

9.  Trace  the  curve 
9  ay'^  =  x{x  —  3  a)^, 

and  find  the  circumference  of  the 
loop.  Ans.  4  a  V3. 

10.  Find  the  length  of  the  curve 
y  —  e^  from  a:  =  0  to  ar  =  f  log  2. 

116.  Surfaces  of  revolution. 
The  surface  generated  by  the 
rotation  of  a  plane  curve 
about  a  line  AB  in  its  plane 
is  easily  expressed  as  a  line 
integral. 
Fig.  71  Let  US  inscribe  in  the  curve 


DEFINITE  INTEGRAL  AS  LIMIT  OF  A  SUM     169 

Q  a  broken  line  of  n  segments  As/ .  Each  of  the  segments 
As/  generates  in  the  rotation  the  frustum  of  a  right  cir- 
cular cone,  the  radii  of  whose  bases  may  be  called  r^  and 
r^  -\-  Ar^-.  The  surface  of  this  conical  frustum  is,  by  ele- 
mentary geometry,  the  circumference  of  the  middle  sec- 
tion multiplied  by  the  slant  height,  or  ^ir^r^  -h  |^Ar^)As/. 

n 

The    sum    of    these    surfaces,    V  2  7r(rj  +  \  Ar^) As/,    ap- 

1=1 
proaches  as  its  limit  the  required  surface  of  revolution. 
Hence,  by  the  theorems  of   §  113  and  §  109,  we  have 

S=  lim  V2'rrnAs/  =2-17      r ds. 

Example:  Find  the  surface  of  a  paraboloid  of  revolu- 
tion bounded  by  a  right  section  through  the  focus. 

Given  the  equation  of  the  generating  parabola  ?/^  =  4  ax^ 
we  have  

=  2  TT  r%\/l  +  ^dx  =1iT  f  V?/2  +  4  ^2  dx 

=  2  TT  I     V4  ax  -\-  ■iaP'dx  = (^ax  +  a^)  ^ 

Jo  a     3  . 

=  |7r«2(2V5-l). 

EXERCISES 

1.  Work  the  above  example,  using  y  as  the  variable  of  integration. 

2.  Find  the  surface  of  a  sphere,  using  polar  coordinates. 

3.  Find  the  surface  of  a  sphere,  using  cartesian  coordinates. 
Evaluate  the  integral  in  various  ways  (cf.  Ex.  10,  p.  146). 

4.  Find  the  surface  generated  by  revolving  the  cubical  parabola 

a^y  =  x^  about  OX,  from  x  =  0  to  x  =  a.  Ans.    ^  (10  VIO  -  1). 

27 

5.  Find   the    surface   generated  by  revolving  (r/)  about  OX,   (h) 

about  OY,  the  arc  of  the  carve  y  =  '    between  the  minimum  point 

6  X  .  ^ 

and  the  point  x  =  2.  A  ns.    (a)  i^;   {h)  ^(15  +  4  log  2). 

16  4 


170 


CALCULUS 


1  x^ 

6.  Trace  the  curve  ?/  =  -  log  x  —  ~  (cf.  §  69),  and  obtain  the  sur- 
face generated  by  revolving  the  curve  about  OY  from  x  =  1  to  x  =  2. 

Arts.    10.47. 

7.  Find     the     surface    generated     by    revolving    the    catenary 

_a     '-       _£ 
y  —  2  (e**  +  e  "),  («)  about  OZ,  (b)  about  OF,  from  a;  =  0  to  a;  =  a. 

Ans.    (b)  27ra2^1--]. 

8.  Find  the  surface  generated  by  the  revolution  of  an  arch  of  the 
cycloid  about  its  base.  Ans.    %*  Tra^. 

9.  Find  the  surface  of  a  torus.  Ans.   Aiir'-ah. 

10.  Find   the   entire   surface   generated   by   revolving   the    curve 
8 a V  =  a:2(a2  _  a;2)  about  OZ.  Ans.^ira^ 

11.  Find  the  surface  formed  by  revolving  the  four-cusped  hypo- 

1  2.  2 

cycloid  x^  +  y^  =  a^  about  OX.  Ans.    '^^ira^. 

12.  Find  the  surface  cut  from  a  sphere  by  a  circular  cone  of  half- 
angle  a  with  its  vertex  at  the  center  of  the  sphere. 

117.  Cylindrical 
surfaces.  Given  a 
cylinder  whose  di- 
rectrix is  a  plane 
curve  (7,  the  area  of 
any  portion  of  the 
cylinder  may  be 
found  as  follows : 
Inscribe  in  (7  a 
broken  line  of  sesf- 
ments  As/,  •••,  As„', 
and  inscribe  in  the 
required  area  a  set 
of  rectangles  of  alti- 
tude A<  and  base  AsJ.  The  limit  of  the  sum  of  these  rect- 
angular areas  is  the  area  on  the  cylinder : 

n  ^ 

S=  lim    VhiLSi'  =  \  hds. 


Fig.  70 


^ 


DEFINITE  INTEGRAL  AS  LIMIT  OF  A  SUM     171 

Example :    Find   the   area   on   the   cylinder  oi^  -\-  z^  =  a^ 
included  between  the  planes  y  =  0^  y  =  mx. 

Denoting  by  O  the  circular  arc  APB,  we  have 


aS'  =  4J  yds  =  -i\    mx^l 

=  4m|    x\l -] — dx 

i         C^xdx 
=  4  ma  I    

Jo     z 

=  —  4  ma  I    (?2  =  4  ma'^. 


Fig.  72 


EXERCISES 

1.  In  the  above  example,  find  the  area  of  the  section  cut  by  the 
plane  y  =  mx. 

2.  Find  the  surface  of  the  cap  in  Ex.  14,  p.  163.      Ans.    128  sq.  in. 

3.  Find  the  surface  of  the  cylinder  x^  +  y^  =a^  included  between 
the  planes  z  =  x,  z  =  3  x. 

4.  Find  the  surface  cut   off  on   the  cylinder  x"^  +  y^  =  a^  by  the 


paraboloid  of  revolution  x'^  -\-  y^  =  hx 


Ans.   2  7r 


5.  Find  the  area,  in  the  first  octant,  of  the  section  of  the  cone 
x^  —  2/^  +  2'^  =1  0  by  the  plane  x  -\-  y  =  a. 

6.  The  center  of  a  sphere  of  radius  2  a  is  on  the  surface  of  a  cyl- 
inder of  radius  a.  Find  the  surface  of  the  cylinder  intercepted  by 
the  sphere.  Ans.    16  a^. 

7.  Find  the  surface  on  the  cylinder  z^  =  Alqx  inside  the  cylinder 
y2  =  4  ax,  from  x  =  0  to  x  =  "d  a.  Ans.   ^^  or. 

8.  Work  the  example  of  §  117,  using  polar  coordinates. 


CHAPTER   XVI 
INTEGRAL  TABLES 

118.  Use  of  tables.  In  the  solution  of  problems  in- 
volving integration,  the  work  may  frequently  be  much 
shortened  by  the  use  of  a  table  of  integrals.  Many  such 
tables  have  been  prepared  ;  the  references  below  are  to 
B.  O.  Peirce's  Short  Table  of  Integrals  (Ginn  &  Co.). 

The  chief  object  in  using  a  table  is  to  save  time.  The 
student  is  therefore  not  making  the  best  use  of  the  table 
unless  he  is  so  familiar  with  its  contents  and  arrange- 
ment that  he  can  tell  at  a  glance  whether  the  desired 
formula  is  likely  to  be  given  and  where  it  is  to  be 
found.  Further,  it  should  be  remembered  that  in  many 
cases  the  result  may  be  found  by  the  methods  of  Chap- 
ter XII  in  less  time  than  would  be  required  if  the  table 
were  used. 


— 

x{\  —  x^ 


;(1  —  aP") 
Let    us    use    formula    bS    of    the    table,    with    a  =  1, 


s, 


dx  1 1  x^ 

lop- 


J ^2  2 

I    a^e^  dx. 
0 

This  integral  is  not  given  explicitly  in  the  table,  but 
it  resembles  formula  402.     Making  the  substitution 

a:2  =  2,  2xdx=  dz^ 


we  find 

j   A'^'^dx  =  \ 

j   ze''dz  =  ^e'(z  —  l) 

' 

172 

4 

—  3. 

~  2 
0 


+h 


INTEGRAL  TABLES  173 


EXERCISES 


Evaluate  the  following  integrals,  using  a  table  whenever  a  saving 
of  time  may  be  effected  by  so  doing. 

1.  C^ — 'I^ .,  Ans.   \  arctan  "  ^  +  ^  +  C. 

2.  j  -  "  "^      zr.  vln.s.  log  ( Vl  +  a;  +  x"-^  +  a:  +  |)  +  C 


Vl  +  a:  +  a;2 


.    c     fix  g  r^_^_^/^^ 

J  (1  +x-)'^*  ■  -^  (1  -2x)2' 

6.     f  xVl  +  ^-r/x.  7.  C ^^^^ . 

J  *^  (1  +  3  xy 

8.     ixy/\  —  xdx.  9.  I  sin'*  a- (/x. 

0.     I 11.  Xxcosx^dx. 

•^  \  +  cos  X  ♦^ 

2.     fx'V'^a:.  13.  ^ xh-^  dx. 

4.     i  arcsin  x  f/ar.  15.     \\o^^xdx. 


6. 


i  arcsin  x  f/ar.  15.     j  log"^ 

C"  __^dx__^  yj      p I 

»/o  ,/.,2  I  ^2  *   ^0  ,/;;2 


//y 


Vft^  +  x'^  *^°  V«^  +  //'^ 

^    j^  V2;  +  t'^dt:  Ans.  Va  -  i  log  (2  +  V3). 

flog  2  .  71- 

>•    J        Ve^  -  1  dx.  Ans.2--^- 

20.  Find  the  area  bounded  by  the  hyperbola  x^  —  if-  =  o?-  and  the 
line  X  =  2a. 

21.  Find  the  length  of  the  arc  of  a  parabola  from  the  vertex  to  the 
end  of  the  latus  rectum.  Ans.  2.29  a. 

22.  Find  the  surface  generated  by  rev^olving  the  curve  y  —  e""  about 
OX  from  a:  =  0  to  :c  =  1. 

23.  Find  the  area  inside  the  four-cusped  hy^Docycloid  x  =  a  cos^  6, 
y  =  a  sin^O.  Ans.  f  Tra^. 

24.  Find  the  area  of  the  ellipse  r= ;:,  where   e   is  the 

^  .  .^  1  +  e  cos^ 

eccentricity. 


174  CALCULUS 

25.  Find  the  volume  generated  by  revolving  one  arch  of  the  cycloid 
about  its  base.  Ans.  5  Tr^a*. 

26.  Find  the  surface  generated  by  revolving  the  curve  3  a^x  +  y^  =  0 
about  OF  from  y  =  0  to  y  =  a. 

27.  Find  the  surface  generated  by  revolving  one  arch  of  the  sine 
curve  about  the  a:-axis.  Ans.  2  7r[V'2  +  log  (1  +  V2)]. 


CHAPTER   XVII 


IMPROPER    INTEGRALS 


119.  Definitions.  Definite  integrals  in  which  either  or 
both  of  the  limits  of  integration  are  infinite,  and  also  those 
in  which  the  integrand  becomes  infinite  within  the  inter- 
val of  integration,  are  called  improper  integrals.  Such  in- 
tegrals have  no  meaning  under  the  definitions  so  far  laid 
down  (see  §§  98, 102)  ;  we  proceed  to  show  how  they  may 
arise,  and  to  find  under  what  conditions  a  meaning  can  be 
assigned  to  them. 

Examples :     (a)  The  area  under  the  curve  y  =  —^  from 


a;  =  1  to  a:  =  6  is  evidently 

•^1    X^  X 


x'- 


When  h  becomes  infinite,  the  area  approaches  the  limit  1. 
This  limit  we  define  as  the  area  "  bounded  "  by  the  curve, 
the  a:-axis,  and  the  line  a:  =  1,  although  it  is  not  properly 
a  bounded  area  in  the  usual  sense  of  the  term.  Symboli- 
cally we  write 

r  dx^_X 
*^\     x^  x_ 

175 


=  1. 


176 


CALCULUS 


The  first  thought  might  be  that  the  area  in  the  figure 
would  increase  indefinitely  as  the  right-hand  boundary 
recedes.     Our  result  shows  that  this  is  not  the  case  —  the 

area  is  always  less  than  1.     . 

(5)   The    area    under    the     curve 
x'lp'  =  1  from  x=  a  (a>  O')  to  a;  =  1  is 


A  =  r^  =  2Vx 

^"  ^x 


=  2-2V^. 


When  a  approaches  0,  the  initial  ordi- 
nate becomes  infinite  ;  the  area  ap- 
proaches the  limit  2.  This  limit  we 
define  as  the  area  in  the  first  quadrant 
"  bounded "  by  the  curve,  the  axes, 
and  the  line  x==  1.  For  brevity,  we 
write  merely 


X 


=  2 


Fig.  74 


'0    ^x 

but  it  must  be  borne  in  mind  that  the 
geometric  interpretation  is  quite  different  from  that  of  the 
ordinary  integral. 

(<?)  Let  us  try  to  find  the  area  under  the  curve  y  —  — 

x^ 

from  x  =  —  1  to  a:  =  1.     If  we  were  to   work    carelessly, 

without   noticing    that    the    integrand    becomes    infinite 

within  the  interval,  we  might  write 

1 


•  -1  X^  X, 


=  -2, 


which  is  obviously  absurd.      But  if  we  write 

dx 


A  —  lim  I       — -  -f  lim    I 

-ll-Vlhnf-iT. 

X^-l        «"^-oL        XJa'* 


lim 


IMPROPER  INTEGRALS  177 

it  is  clear  that  the  limits  do  not  exist,  and  the  area-integral 
has  no  meaning. 

These  examples  suggest  the  following  definitions : 

f  (x)  dx  =  lim    I  f(x)  dx  ; 

I     f{x}dx=  lim    I    f{x^dx^ 

provided  the  limits  exist. 

2.  If  fix)  becomes  infinite  as  x-^h~^ 

\  f{x)dx=  lim    I    f{x}dx; 
\if(x)  becomes  infinite  as  2:->a+, 

fQc)dx=\\m    I    f{x)dx^ 

provided  the  limits  exist. 

3.  If /(a:)  becomes  infinite  as  a;->c,  where  a  <  c  <  b, 

Jrb  re'  .       rb 

I    f(x)dx=    lim     I    f\x^dx-{-  lim    I    /{x^dx^ 

provided  the  limits  exist. 

120.  Geometric  interpretation.  It  is  obvious  that  an  in- 
tegral with  an  infinite  limit  may  be  interpreted  in  general 
as  the  area  under  a  curve  which  is  asymptotic  to  the  2:-axis  ; 
an  integral  whose  integrand  becomes  infinite  may  usually 
be  thought  of  as  the  area  between  a  curve  and  a  vertical 
asymptote.  Of  course,  as  in  example  (c)  of  §  119,  these 
integrals  may  not  have  any  meaning  in  a  given  case. 

EXERCISES 

Evaluate  the  following  integrals. 

1-   («)     C~'^     Q^)    T— 5     (0    Tcosxr/x;  (^0    C  e^dx. 

Ans.   (a)  i;  (Jj)  meaningless;  (c)  meaningless;  (d)  ^. 

2.    \     —    '  Ans.    TT.      3.     i    ^-         Ans.    Meaningless. 

^-Wl-x^  ^0    X 

4.    r  —  Ans.    2.       5.     C"       ^^       •       Ans.    2\/2^. 

^  ^  Vx  ^^    V2a-t 

N 


178  CALCULUS 

6.  Trace  the  curve  y  =  — — ,  and  find  the  area  between  the 

curve  and  its  asymptote.  Ans.   4:7ra^. 

7.  Find  the  volume  generated  by  revolving  the  area  of  Ex.  6 
about  the  asymptote.  Ans.   47r%^. 

8.  Find  the  area  in  the  fourth  quadrant  bounded  by  the  curve 
xy^  =  (x  —  1)'-^  and  the  coordinate  axes. 

9.  Find  the  area  in  the  second  quadrant  under  the  curve  y  =  e''. 

10.  Find    the    volume    generated    by   revolving    (a)    about    OX, 
(b)  about  OY,  the.  area  in  the  second  quadrant  under  the  curve  y  =  e'. 

Ans.    (a)  |;  (6)  2  7r. 

11.  The  area  in  example   (6),  §  119,  revolves   about  the  ?/-axis. 
Find  the  volume  generated. 

12.  Find  the  surface  generated  by  revolving  about  OX  that  portion 
of  the  curve  y  =  e"^  which  lies  to  the  left  of  the  ^-axis.       Ans.   2.29  tt. 

13.  Trace  the  curve  x(x  —  yY  =  a^,  and  find  the  area  bounded  by 
the  curve,  the  ?/-axis,  and  the  line  a;  =  4  a.  Ans.  2  a^. 

14.  Find  the  volume  generated  by  revolving  about  OY  the  area 
under  the  curve  y  =  e~2=^\     Check  by  solving  in  tvv^o  ways. 


CHAPTER   XVIII 

CENTROIDS.    MOMENTS   OF   INERTIA 

I.    Centroids 

121.  Mass;  density.  The  student  is  assumed  to  be 
familiar  with  the  idea  of  mass  as  introduced  in  physics. 

A  mass  is  said  to  be  homogeneous  if  the  masses  contained 
in  any  two  equal  volumes  are  equal.  In  all  other  cases 
the  mass  is  heterogeneous.  In  the  present  chapter  we  con- 
fine our  attention  to  homogeneous  masses. 

The  density  S  of  a  homogeneous  mass  is  the  ratio  of  the 
mass  Mio  the  volume  l^that  it  occupies  : 

V 

That  is,  the  density  is  the  mass  per  unit  volume. 

Although  every  physical  mass  occupies  a  certain  volume 
or  three-dimensional  portion  of  space,  nevertheless  it  is 
frequently  desirable  to  introduce  the  idea  of  the  material 
particle^  or  geometric  point  endowed  with  mass.*  The 
mass-point  may  be  imagined  as  the  limiting  form  ap- 
proached by  a  body  whose  dimensions  approach  0,  while  the 
density  increases  in  such  a  way  that  the  mass  remains 
finite. 

Similarly  we  may  think  of  masses  of  one  dimension  and 
of  two  dimensions  —  i.e.  of  material  curves  and  surfaces. 
Such  masses  are  represented  approximately,  for  example, 
by  slender  wires  and  thin  sheets  of  metal.  In  these  cases 
we  define  the  density  as  "linear  density,"  or  mass  per 
unit  length, 

S 

*  This  notion  is  fundamental  in  studying  the  motion  of  a  rigid  body. 

179 


180  CALCULUS 

and  "surface  density,"  or  mass  per  unit  area, 

respectively. 

122.  Moment  of  mass.  The  product  of  a  mass  m,  con- 
centrated at  a  point  P,  by  the  distance  Z  of  P  from  a  given 
point,  line,  or  plane,  is  called  the  moment*  of  m  with  re- 
spect to  the  point,  line,  or  plane.     Denoting  this  moment 

by  (x,  we  have 

Q  —  ml. 

If  a  system  of  points  P^  P^^  •••,  P„,  having  masses  Wj, 
jTZg,  •••,  m^  respectively,  be  referred  to  cartesian  coordinate 
axes,  the  moments  of  the  system  with  respect  to  the  three 
coordinate  planes  are  respectively 


n 


^1 
n 


In  case  the  particles  all  lie  in  one  of  the  coordinate 
planes,  the  moments  with  respect  to  coordinate  planes 
reduce  to  moments  with  respect  to  coordinate  axes. 

The  idea  of  mass-moment  may  be  extended  to  the  case 
of  a  continuous  mass  by  thinking  of 'the  mass  as  composed 
of  an  indefinitely  large  number  of  particles.  A  precise 
definition  will  be  laid  down  in  §  187.  The  actual  compu- 
tation of  such  a  moment  is  usually  effected  by  means  of 
definite  integrals;   we  return  to  this  question  presently. 

123.  Centroid.  Given  any  mass  if,  let  G-y^^  G-^^  G-^y  de- 
note the  moments  of  the  mass  with  respect  to  the  coordi- 
nate planes.  The  point  C  whose  coordinates  ^,  y,  ^,  are 
given  by  the  formulas 

Mx  =  a^,.   My  =  a,,,   M-z=a,y 

*  More  precisely,  the  simple  moment^  or  moment  of  first  order. 


CENTROIDS.     MOMENTS  OF  INERTIA  181 

clearly  has  the  property  that  the  moment  of  the  mass  with 
respect  to  each  of  the  coordinate  planes  is  the  same  as  if 
the  whole  mass  were  concentrated  at  that  point. 

It  is  easily  shown  that  this  property  holds  for  moments 
with  respect  to  any  other  plane.  The  proof  for  the  gen- 
eral case  requires  the  use  of  multiple  integrals  (Chapter 
XXIII);  for  a  system  of  mass-particles  the  proof  is  as 
follows.     Let 

(1)  ax  -\-  hy  -\-  cz  =  p 

be  the  equation  of  any  plane  in  the  normal  form  ;  let 
^,  ^j,  p^,  '"'>  Pn  be  the  distances  of  the  points  O,  P^  P^^  •••, 
P^  from  this  plane.     Now 

p^  =  ax^  -f  %i  +  cz^  —  p, 

p^=  ax^  +  hyr,  +  cz^  -  p, 
so  that 

n  n  n  n  n 

t=l  J=l  i=l  ?=1  1=1 

=  aMx  -h  hMy  +  c3Iz  -  Mp 
=  M(^ax  -{-  by  -\-  ez  —  p} 
=  Mp. 
That  is,  the  moment  of  the  system  with  respect  to  the 
plane  (1)  is  the  same  as  if  the  whole  mass  were  concen- 
trated at  O, 

The  point  O  is  called  the  center  of  mass^  or  centroid  *  ; 
The  centroid  of  a  mass  is  a  point  such  that  the  moment 
of  the  mass  ivith  respect  to  any  plane  is  the  same  as  if  the 
ivhole  mass  ivere  concentrated  at  that  point. 

By  §  122,  the  coordinates  of  the  centroid  of  a  system  of 
particles  are  given  by  the  formulas 

n  71  .  n 

i=\  i=l  i=l 

*  The  centroid  coincides  with  the  center  of  gravity^  and  is  frequently 
so-called  ;  but  the  term  centroid  is  in  some  respects  preferable. 


182  CALCULUS 

In  the  actual  determination  of  centroids,  the  following 
considerations  are  often  useful  (the  first  two  apply  only  to 
homogeneous  masses): 

(a)  If  the  body  has  a  geometrical  center,  that  point  is 
the  centroid. 

(5)  Any  plane  or  line  of  symmetry  naust  contain  the 
centroid. 

(c)  If  the  body  consists  of  several  portions  for  each  of 
which  the  centroid  can  be  found,  each  portion  may  be 
imagined  concentrated  at  its  centroid:  the  problem  thus 
reduces  to  the  consideration  of  a  set  of  particles. 

124.  Centroids  of  geometrical  figures.  It  is  clear  that, 
for  a  homogeneous  body  of  given  size  and  shape,  both  the 
mass  and  its  moment  with  respect  to  any  plane  are  pro- 
portional to  the  density  h.  Hence,  in  the  formulas  for  ^, 
y^  i,  8  cancels  out  from  both  members,  leaving  the  co- 
ordinates of  the  centroid  independent  of  the  density. 
We  may  therefore  without  loss  of  generality  take  S=l, 
and  are  thus  led  to  speak  of  centroids  of  geometrical 
figures  —  volumes,  areas,  and  lines  —  without  reference  to 
the  idea  of  mass. 

EXERCISES 

1.  Find  the  centroid  of  the  following  plane  systems  of  particles : 
(a)  Equal  particles  at  (0,  0),  (4,  2),  (3,  -  5),  (-  2,  -  3). 

(h)  A  mass  of  2  units  at  (0,  1),  one  of  3  units  at  (3,  —  3),  one 
of  6  units  at  (4,  1). 

2.  Four  particles  of  mass  2,  4,  6,  8  units  are  placed  at  the  points 
(0,  0,  0),  (0,  2,  2),  (4,  1,  5),  (-  3,  2,  -  1)  respectively.  Find  the 
centroid. 

3.  Show  that  the  centroid  of  two  particles  divides  the  line  joining 
them  into  segments  inversely  proportional  to  the  masses. 

4.  Show  that  the  centroid  of  three  equal  particles  lies  at  the  in- 
tersection of  the  medians  of  the  triangle  having  the  three  points  as 
vertices. 

5.  Equal  particles  are  placed  at  five  of  the  six  vertices  of  a  regular 
hexagon.     Find  the  centroid. 


CENTROIDS.     MOMENTS  OF  INERTIA 


183 


6.  Particles  of  mass  1,  2,  •••,  8  units  are  placed  at  the  successive 
vertices  of  a  regular  octagon.     Find  the  centroid. 

7.  Find  the  centroid  of  the  cross  section  of  an  angle-iron,  the  sides 
being  5  in.  and  8  in.,  and  the  thickness  of  each  flange  1  in. 

Ans.   (-V,  f). 

8.  Find  the  centroid  of  the  T-iron  section  (a)  of  Fig.  75,  (b)  of 
Fig.  76. 

I  9.    Find  the  centroid  of  a  wire  frame  in  the  shape  of  the  perimeter 

(a)  of  Fig.  75,  (ft)  of  Fig.  76. 


\' 

<-2'-U 

1" 

n" 

^ 

/ 

— I 

^3^ 

_^ 

<-3'^ 

<-2^ 

4" 


—  6'' 

Fig.  75 


-8^ 


Fig.  76 


10.  From  a  circular  disk  a  round  hole  is  punched  out,  the  two 
circles  being  tangent  internally.  Find  the  centroid  of  the  remaining 
figure. 

11.  From  a  circular  plate  of  radius  4  in.  a  hole  2  in.  square  is  cut 
out,  one  corner  of  the  square  being  at  the  center  of  the  plate.  Find 
the  centroid  of  the  remainder. 

12.  Find  the  centroid  of  a  cylindrical  basin  of  radius  4  in.  and 
depth  3  in.,  if  the  bottom  is  twice  as  thick  as  the  sides. 

13.  A  monument  is  composed  of  a  block  of  stone  of  base  4  ft.  by 
3  ft.  and  height  2  ft.  6  in.,  surmounted  by  a  cube  of  side  2  ft.,  this  in 
turn  supporting  a  sphere  of  radius  1  ft.  Find  the  centroid  of  the 
whole  figure. 

125.    Determination   of   centroids    by   integration.       To 

find  the  centroid  of  a  continuous  mass,  we  must  in  gen- 
eral resort  to  integration.  In  the  most  general  case  mul- 
tiple integrals  (see  Chapter  XXIII)  must  be  used,  but  in 
most  cases  of  practical  importance  the  result  may  be 
obtained  by  a  single  integration. 

In  the  following  discussion  we  restrict  ourselves  to  one-, 
two-,  or  three-dimensional  bodies  of  the  forms  considered 


184  CALCULUS 

in  Chapter  JXV.*  Let  us  choose,  as  in  that  chapter,  a  suit- 
able geometrical  element  (of  volume,  area,  or  length),  and 
denote  the  mass  contained  in  this  element  by  Am^.  Let  x^^ 
yi,  Zi  be  the  coordinates  of  the  centroid-f  of  Aw,.     Then  the 

n 

sum  ^XiArrii  represents  approximately  the  moment  of  the 

1=1 
mass  with  respect  to  the  ^2-plane  (or  the  «/-axis,  in  the 

case  of  a  plane  mass  in  the  a^^-plane),  and  the  limit  of 

this  sum  as  n  becomes  infinitJ^  is  exactly  the  moment  in 

question.     In  this  way  we  obtain  the  following  formulas 

for  the  coordinates  of  the  centroid  : 

M'x  =  lim  Va^Awf, 

n 

M'^  =  lim  ^i/Anii, 

n 

Mz  =  lim  ^zAnii. 

Now  upon  recalling  the  meaning  of  Aw^,  we  see  that 
in  any  given  case  the  above  limits  may  be  expressed  as 
definite  integrals  by  the  fundamental  theorem  of  §  104. 
The  result  may  be  written  in  the  following  form : 

(1)  Mx=  ( X  dm,   My=  iydm,  Mz=  (zdm, 

where  x,  «/,  z  are  the  coordinates  J  of  the  centroid  of  the 
mass-element.  In  any  problem  each  integrand  must  be 
expressed  in  terms  of  a  single  variable,  and  limits  are  to 
be  assigned  in  such  a  way  as  to  extend  the  integration 
over  the  whole  mass. 

In  the  next  few  articles  we  explain  more  in  detail  tlie 
application  of  the  above  formulas. 

*  Nevertheless  the  formulas  obtained  are  applicable,  with  proper 
interpretation,  to  all  masses,  with  no  restriction  whatever. 

t  It  follows  from  the  theorem  of  §  109  that  in  the  expressions  for 
these  coordinates,  all  infinitesimals  may  be  neglected. 

I  Apart  from  infinitesimals. 


» 


CENTROIDS.       MOMENTS  OF  INERTIA 


185 


126.    Centroids  of  plane  areas.*     To  find  the  centroid  of 
a  plane  area  (thin  sheet  or  plate)  we  choose  an  element  of 
area  as  in  §  105  (or  §  106,  if  polar  coor- 
dinates are  used),  and   find  the  centroid 
.    from  the  formulas 


Ax  =  Cx  dA,  Ay  =  i  y  dA, 


*'  where  x  and  y  are  the  coordinates  of  the 

centroid  of  the  element. 

Uxample :  Find  the  centroid  of  the  area 

in    the    first    quadrant    bounded    by    the 

parabola  y^=  -iax  and  its  latus  rectum. 
It      With  the  element  of  area  chosen  as  in 

Fig.  61,  we  have 

^  =  I    y  dx  =  ^  a^ ; 


Fig.  61 


Ax—  I     xydx-=^^a\    . 

Jo  Jo 

Ay  ==  j    U  '  ydx  —  2ai 


x^  dx=  i  a^ 


dx 


a^. 


Hence 


^  =  f«'  «/  =  !«• 


We  may  also  find  y  very  easily  by  taking  the  element 
parallel  to  OX.     Thus 

-^y  —  I      y(.^  ~  ^^dy  =  2  a  j    (a  —  x^dx  —  aP^  etc. 

Jo  Jo 


EXERCISES 

Find  the  centroid  of  the  following  areas.  In  each  case,  draw  a 
figure  and  estimate  the  coordinates  of  the  centroid,  thus  obtaining 
a  rough  check  on  the  result. 

1.  An  isosceles  triangle. 

2.  A  semicircular  area.     Evaluate  the  integral  in  two  ways. 

*  The  problem  of  this  article  is  of  particular  importance  in  the  theory 
of  the  flexure  or  bending  of  beams. 


186  CALCULUS 

3.  One  quadrant  of  an  ellipse,  using  (a)  the  equation  —  +  ^  =  1 

(6)  the  equations  x  =  a  cos  cf>,  y  =  b  sin  <fi.  Ans.    {-^ — ,  ^^ — ). 

4.  Any  triangle.  Ans.    At  the  intersection  of  the  medians. 

5.  Half  an  arch  of  the  sine  curve.  Ans.    ( 1,  -  )• 

6.  The  area  between  the  curves  2  7/  =  x'^,  y  =  x^.     Get  each  coor- 
dinate in  two  ways. 

7.  A  circular  sector.  Ans.   x  =-  a  ^^^"- 

3         a 

8.  One  arch  of  the  cycloid.  Ans.    (tto,  |  a). 

9.  A  semicircular  area,  using  polar  coordinates. 

10.  A  circular  segment.     Check  by  putting  h  =  a. 

11.  A  trapezoid. 

12.  One  loop  of  the  curve  r  =  a  cos  2  0. 

13.  The  area  under  the  curve  y  —  e"^  from  a:  =  0  to  a:  =  1. 

14.  The  area  bounded  by  the  curve  y  =     ^     ,  the  a:-axis,  and  the 
maximum  ordinate. 

15.  The  area  bounded  by  the  parabola  y  =  x^,  the  a;-axis,  and  the 
line  X  =  3. 

16.  The  area  bounded  by  the  curves  y  =  x,  y  =  2  x,  y  =  x^. 

17.  The  area  bounded  by  the  catenary  y  =  ^[e^  ■}-  e~'^j  ,  the  axes, 
and  the  line  x  =  a. 

18.  The   area   swept   out   by  the   radius   vector  of   the   spiral  of 
Archimedes  r  =  aO  in  the  first  revolution. 

19.  Half  the  area  of  the  cardioid  r  =  a(l  —  cos  0). 

20.  The  upper  half  of  the  loop  of  the  curve  ay^  =  ax^  —  x^. 

21.  The  area  of  Ex.  7,  p.  153. 

22.  From  one  corner  of  a  square  of  side  a,  a  triangle  of  sides  I  a, 
^  a  is  cut  off.     Find  the  centroid  of  the  remaining  area. 

23.  From  a  circle  an  inscribed  isosceles  right  triangle  is  cut  out. 
Find  the  centroid  of  the  remainder. 

24.  Prove  the  second  proposition  of  Pappus  : 

The  volume  of  any  solid  of  revolution  is  equal  to  the  product  of 
the  generating  area  into  the  circumference  of  the  circle  described  by 
the  centroid  of  the  area. 


f 


CENTROIDS.     MOMENTS  OF  INERTIA  187 

Solve  the  following  by  using  the  second  proposition  of  Pappus. 

25.  Find  the  volume  of  a  torus.  Ans.   2  7r'^a% 

26.  Find  the  centroid  of  a  right  triangle. 

27.  Find  the  centroid  of  a  semicircular  area. 

•  127.  Centroids  of  volumes.  The  centroid  of  a  volume 
of  revolution  evidently  lies  on  the  axis  of  revolution,  so 
that  a  single  coordinate  determines  its  position.  Taking 
the  axis  of  revolution  as  axis  of  x,  we  have 

Vx=CxdV, 

where  the  element  of  volume  is  chosen  as  in  §  107  or 
§  108,  and  where  x  is  the  :c-coordinate  of  the  centroid  of 
the  element. 

In  certain  special  cases  the  centroids  of  other  solids 
may  be  found  by  a  simple  integration,  but  in  general  we 
must  resort  to  multiple  integrals. 

EXERCISES 

Find  the  centroid  of  the  following  volumes.  Draw  a  figure  in 
each  case  and  estimate  the  coordinates  of  the  centroid. 

1.  A  hemisphere.     Solve  in  two  ways.  Atis.    x  =  ^a. 

2.  A  right  circular  cone.  Ans.   x  =  I  h. 

3.  A  paraboloid  of  revolution  bounded  by  a  right  section  through 
the  focus. 

4.  Half  an  ellipsoid  of  revolution.     Solve  in  various  ways. 

6.    A  spherical  segment  of  height  h.     Check  by  putting  h  =  r. 

6.  The  volume  generated  by  revolving  (a)  about  OX,  (6)  about 
OY,  the  area  under  the  curve  y  =  e""  from  a:  =  0  to  a;  =  1. 

7.  The  volume  generated  by  revolving  half  an  arch  of  the  cycloid 
about  its  base. 

8.  The  volume  formed  by  rotating  the  area  under  the  parabola 
2/2  =  4  aa;  from  x  =  0  to  x  =  a,  (a)  about  the  y-axis ;  (b)  about  the 
latus  rectum ;  (c)  about  the  line  ?/  =  2  a. 

9.  The  volume  in  Ex.  1.3,  p.  163.  Ans.    Q  5,  f  a,  i  a). 

10.  The  volume  in  Fig.  68.  Check  all  three  coordinates  by  solving 
again  with  the  element  chosen  in  a  different  way. 

11.  An  elliptic  cone.  Ans.   x  =  ^h. 


188  CALCULUS 

12.  One  quarter  of  a  right  circular  conoid  (see  Ex.  12,  p.  163). 

13.  One  quadrant  of  the  banister  cap  in  Ex.  14,  p.  163. 

14.  A  tetrahedron  three  of  whose  faces  are  mutually  perpendicular. 

Ans.    (la,  lb,  ^c). 

15.  The  volume  of  Ex.  16,  p.  163.     Solve  in  two  ways. 

16.  One  quarter  of  a  right  pyramid  with  a  square  base. 

17.  The  volume  of  Ex.  18,  p.  163. 

18.  The  volume  of  Ex.  19,  p.  161. 

19.  The  volumes  of  Ex.  6,  p.  160. 

20.  The  volume   formed   by  revolving  about  the  ?/-axis  the  area 

under  the  curve  ?/  =  e    ^     ^     Take  as  the  element  a  cylindrical  shell, 
and  evaluate  the  integrals  in  two  ways. 

21.  Obtain  formulas  for  the  coordinates  of  the  centroid  of  a  wedge 
cut  from  any  solid  of  revolution  by  two  planes  through  the  axis. 

128.  Centroids  of  lines.  In  the  case  of  an  arc  of  a 
plane  curve,  the  fundamental  limits  of  §  125  may  be 
expressed  at  once  as  line  integrals,  by  the  theorem  of 
S113: 


sx  =  I  X  ds^    sy  =  \  y  ds^ 


where  s  is  the  length  of  the  arc  O. 

Example:  Find  the  centroid  of  a  semicircular  wire. 
Taking  the  bounding  diameter  as  axis  of  y,  we  have 


^n  */0         ^  \dnr. 


dx 


==2rxJl+^dx=2ar 


dxj 

X  dx 


r  -'^  -^d^-x' 


Hence 


=  —  2  a^a^  —  x^ 
Jo 

2  ^2      2  «2       2a 


2a\ 


X  = 


s         ira        IT 
By  symmetry,  ^  =  0. 


CENTROIDS.     MOMENTS  OF  INERTIA  189 

129.  Centroids  of  curved  surfaces.  The  coordinates  of 
the  centroid  of  a  surface  of  revolution  (§  116),  or  of  a 
cylindrical  surface  (§  117),  may  be  expressed  in  terms  of 
line  integrals. 

The  required  integrals  are  easily  built  up  in  each 
problem. 

EXERCISES 

1.  In  the  example  of  §  128,  evaluate  the  integral  by  expressing 
the  integrand  in  terms  of  y. 

Find  the  centroid  of  each  of  the  following  figures.  • 

2.  A  semicircular  wire,  using  polar  coordinates. 

3.  The  arc  of  the  curve  ay^  =  x^,  from  x-  =  0  to  a:  =  5  a. 

X  X 

4.  The  arc  of  the  catenary  ?/=-(e«+e    ")  between  two  sym- 
metric  points. 

5.  The  arc  of  the  semicycloid  (from  cusp  to  vertex) . 

Ans.    (^a,^a). 

6.  A   hemispherical   surface,  using    (rt)  cartesian,  (h)  polar  co- 
ordinates. .4ns.    X  =\a. 

7.  The  lateral  surface  of  a  right  circular  cone.  Ans.   Ic  =\h. 

8.  The  total  surface  of  a  right  circular  cone. 

9.  The  cylindrical  surface  in  Fig.  72. 

10.  The  area  in  Ex.  5,  p.  171. 

11.  The  surface  in  Ex.  7,  p.  171. 

12.  The  surface  of  a  paraboloid  of  revolution  bounded  by  a  right 
section  through  the  focus. 

13.  Prove  the  Jirst  proposition  of  Pappus : 

The  surface  of  a  solid  of  revolution  is  equal  to  the  length  of  the 
generating  arc  multiplied  by  the  circumference  of  the  circle  described 
by  the  centroid  of  the  arc. 

14.  Find  the  surface  of  a  torus.  Ans.    i-Tr'-ab. 

15.  Find  the  centroid  of  a  semicircular  wire  by  tlie  first  propo- 
sition of  Pappus. 


190  CALCULUS 

II.    Moments  of  Inertia 

130.  Moment  of  inertia.  The  product  of  a  mass  w,  con- 
centrated at  a  point  P,  b}^  the  square  of  the  distance  r  of 
P  from  a  fixed  line,  or  axu^  is  called  the  moment  of  the 
second  order,  or  moment  of  inertia^  of  m  with  respect  to 
the  given  axis  : 

/=  mr^. 

The  moment  of  inertia  of  a  system  of  such  masses  is  of 
course  the  sum 

n 

rriiri 


i=X 


2 

t'  i  ' 


If  we  think  of  a  "  continuous  "  mass  as  composed  ulti- 
mately of  particles,  the  meaning  of  moment  of  inertia  of 
such  a  mass  becomes  clear.  An  analytic  definition  will  be 
given  in  §  187. 

The  moment  of  inertia  of  a  homogeneous  body  is  pro- 
portional to  the  density.  Taking  5  =  1,  we  may  speak  of 
"moment  of  inertia"  of  areas,  volumes,  etc.,  no  idea  of 
mass  being  involved. 

131.  Radius  of  gyration.  The  moment  of  inertia  of  any 
mass  M  may  always  be  written  in  the  form 

where  the  quantity  R  is  called  the  radius  of  gyration^  or 
radius  of  inertia^  of  M  with  respect  to  the  given  axis  of 
moments.  The  meaning  of  the  radius  of  gyration  is 
obvious :  it  is  the  distance  from  the  axis  at  which  a  parti- 
cle of  mass  M  must  be  placed  in  order  to  have  the  same 
moment  of  inertia  as  the  original  mass. 

132.  Determination  of  moment  of  inertia  by  integration. 

The  actual  computation  of  the  moment  of  the  second  order 
of  a  continuous  mass  is  effected  by  integration  in  much 
the  same  way  that  the  moment  of  the  first  order  (§  125) 
is  determined.     In  the  general  case,  we    must   have  re- 


CENTROIDS.     MOMENTS  OF  INERTIA  191 

course  to  double  or  triple  integrals  (Chapter  XXIII)  ;  but 
for  the  simple  cases  of  practical  importance  the  result  can 
usually  be  found  by  a  single  integration. 

For  the  present  we  consider  only  such  bodies  as  were 
studied  in  Chapter  XV.  Let  us  choose  a  geometrical  ele- 
ment (of  volume,  area,  or  length)  in  some  suitable  way, 
and  denote  the  mass  of  thi's  element  by  Am^.  The  element 
must  he  so  chosen  that  its  radius  of  gyration  is  known  *  ;  let 
Ti  denote  this  radius.     Then  the  sum 

n 

is  approximately  the  moment  of  inertia  of  the  mass  with 
respect  to  the  given  axis,  and  the  limit  of  this  sum  is 
exactly  the  required  moment : 

n 

Z=  lim  Z/  r^^m^. 

«— ^OO     1=1  * 

Now  if  we  apply  the  fundamental  theorem  of  §  104,  the 
above  limit  appears  as  the  definite  integral 


(1)  /  =ff- 


dm. 


where  r  is  the  radius  of  gyration  of  the  mass  element  with 
respect  to  the  axis  of  moments.  Of  course  the  integrand 
must  be  expressed  in  terms  of  a  single  variable  and  the 
integration  must  be  extended  over  the  whole  mass. 

Just  as,  in  finding  centroids,  we  must  take  an  element 
the  position  of  whose  centroid  is  known,  so  here  the 
essential  point  is  to  choose  an  element  whose  radius  of 
gyration  is  known.  Thus  the  moment  of  inertia  of  a 
plane  area  (or  of  a  thin  sheet  of  mass)  with  respect  to  a 
line  in  its  plane  may  be  found  by  taking  as  the  element 
a  rectangle  with   its    finite    side  parallel  to  the  axis  of 

*  In  the  expression  for  the  radius  of  gyration,  infinitesimals  may  as 
usual  be  neglected. 


192 


CALCULUS 


moments,  since  then  the  radius  of  gyration  of  the  element 
is  simply  its  distance  from  the  axis. 

To  find  the  moment  of  inertia  of  a  volume  of  revolution 
about  the  axis  of  revolution  it  is  usually  best  to  choose 
the  element  as  in  §  108. 

Examples :  (a)  Find  the  moment  of  inertia,  with  re- 
spect to  the  ?/-axis,  of  the  area  bounded  by  the  parabola 
2/2  _.  4  ^2;,  the  ic-axis,  and  the  latus  rectum. 

Taking  the  element  parallel  to  OY^  we  find 

Iy=  \  ""x^y  dx  =  2  Va  I   x^  dx  =  |  a^. 
Since  the  mass,  or  area,  is 


we  may  write 


ly  =  f  Ma\ 
which  shows  that  the  square  of  the  radius 
of  gyration  is 


i22=  3 


a^ 


(5)  The  area  in  Fig.  61  revolves  about 
the  ?/-axis.  Find  the  moment  of  inertia 
of  the  volume  generated,  with  respect  to 
the  axis  of  revolution. 

Take  as  element  of  volume  the  cylin- 
drical shell  generated  by  the  rectangle 
shown  in  Fig.  61,  so  that 

c?  F^  =  2  iTxy  dx. 
The  radius  of  gyration  of  this  shell  about 
the  ?/-axis  is  evidently  x.     Hence 

Iy=  2tt  \    0^  ■  xy  dx  =  4:  jr^a  I    a;^  dx 

=  -|  ira^. 
The  mass,  or  volume,  is 

M=  2  7r  i    xy  dx  =  ^  ira^ 
whence  ly  =  |  Ma^, 


Fig.  61 


CENTROIDS.     MOMENTS  OF  INERTIA  193 

EXERCISES 

Find  the  following  moments  of  inertia. 

1.  A  particle  of  mass  3  units  at  (0,  0),  one  of  4  units  at  (2,  2), 
and  one  of  5  units  at  (—  1,  —  3),  with  respect  to  each  of  the  coordi- 
nate axes. 

2.  Equal  particles  at  (0,  0,  0),  (0,  5,  0),  (3,  4,  3),  with  respect  to 
each  coordinate  axis. 

3.  Equal  particles  at  each  corner  of  a  cube,  («)  wdth  respect  to  an 
edge  of  the  cube,  (6)  with  respect  to  a  diagonal  of  one  face. 

.4  ns.  (a)  Ma^;  (h)  lMa\ 

4.  A  straight  rod  or  wire  with  respect  to  a  perpendicular  through 
one  end.  Ans.   \  Ml'^. 

5.  A  rectangle  about  one  side.  Ans.   i  Ma"^. 

6.  A  circular  disk  with  respect  to  a  diameter.  Ans.    \Mcfi. 

7.  The  area  in  the  example  of  §  132  with  respect  to  the  a:-axis, 
(a)  taking  the  element  parallel  to  OX,  (h)  taking  the  element 
parallel  to  OF  and  using  the  result  of  Ex.  5. 

8.  (a)  An  isosceles  triangle,  {h)  any  triangle,  with  respect  to  the 
base. 

9.  A  circular  disk  of  radius  4  in.  with  a  square  of  side  2  in.  cut 
out  of  the  center,  with  respect  to  a  diameter  parallel  to  a  side  of  the 
square. 

10.  An  ellipse  with  respect  to  each  of  its  axes,  using  (a)  the  carte- 
sian equation,  (b)  the  equations  x  =  a  cos  <f),  y  =  b  sin  <f>. 

11.  The  area  bounded  by  the  parabola  3/^  =  4  ax,  the  ?/-axis,  and 
the  line  y  =  2a,  with  respect  to  each  coordinate  axis. 

12.  The  area  in  Fig.  75,  p.  183,  (a)  with  respect  to  the  base, 
(h)  with  respect  to  the  line  of  symmetry. 

13.  The  area  in  Fig.  76,  with  respect  to  the  base. 

14.  A  sphere  with  respect  to  a  diameter.  Ans.  f  Ma^. 

15.  A  cylinder  of  revolution,  with  respect  to  its  axis.  Ans.  |  Ma^. 

16.  A  right  circular  cone  with  respect  to  its  axis.  Ajis.  ^q  Ma^. 

17.  A  paraboloid  of  revolution  bounded  by  the  right  section 
through  the  focus,  with  respect  to  the  axis. 

18.  An  ellipsoid  generated  by  revolving  an  ellipse  about  its  major 
axis,  with  respect  to  the  axis  of  revolution.  Use  («)  the  cartesian 
equation  of  the  ellipse,  (h)  the  equations  x  =  a  cos  <^,  y  =  h  sin  <f}. 


194  CALCULUS 

19.  The  volume  formed  by  revolving  the  area  of  Fig.  61  about  the 
latus  rectum,  with  respect  to  the  axis  of  revolution. 

20.  A  circular  disk  about  its  axis  —  i.e.  the  line  through  the 
center  of  the  disk  perpendicular  to  its  plane.  Ans.    \Ma^. 

21.  A  wire  bent  in  the  form  of  a  square,  with  respect  to  (a)  a  side, 
(6)  a  diagonal. 

22.  A  circular  wire  with  respect  to  a  diameter,  using  (a)  polar, 
(h)  cartesian  coordinates. 

23.  The  arc  of  the  curve  ay"^  =  x^,  from  x  =  0  to  x  =  5  a,  with 
respect  to  the  ?/-axis. 

24.  A  spherical  surface  about  a  diameter,  using  (a)  polar,  (6) 
cartesian  coordinates. 

25.  The  lateral  surface  of  a  cone  of  revolution,  about  its  axis. 

Ans.  ^  MaK 

26.  A  torus,  with  respect  to  its  axis. 

27.  The  surface  of  a  torus,  about  its  axis. 

28.  The  surface  of  a  paraboloid  of  revolution  bounded  by  a  right 
section  through  the  focus,  with  respect  to  the  axis. 

29.  The  surface  in  Ex.  5,  p.  171,  about  tlie  2-axis. 

30.  The  surface  in  Ex.  7,  p.  171,  about  the  ?/-axis. 

31.  The  arc  of  the  curve  ay^  =  x^  from  x  =  0  to  x  =  a  revolves 
about  OY.  Find  the  moment  of  inertia  of  the  surface  generated, 
with  respect  to  the  y-axis. 

32.  Find  the  moment  of  inertia  of  the  volumes  in  Exs.  14,  16,  17, 
18,  by  using  the  result  of  Ex.  20. 

33.  The  area  under  the  curve  y  =  e~^^  revolves  about  the  ?/-axis. 
Find  the  moment  of  inertia  of  the  volume  generated,  with  respect  to 
the  2/-axis.  -  Ans.  2  M. 

133.  Moment  of  inertia  with  respect  to  a  plane.  In  most 
applications  we  are  concerned  with  moment  of  inertia  with 
respect  to  a  line.  Nevertheless  it  is  frequently  useful,  as 
we  shall  see  presently,  to  introduce  the  idea  of  moment 
of  inertia  with  respect  to  a  plane.  The  definitions  and 
discussion  of  §§  130-132  hold  at  once  ior  the  moment  of 
the  second  order  with  respect  to  a  plane  if  we  replace  the 
word  ''  line  "  (or  ''  axis  ")  throughout  by  the  word  "  plane." 


i 


CENTROIDS.     MOMENTS  OF  INERTIA  195 

Example :  Find  the  moment  of  inertia  of  a  sphere  with 
respect  to  a  diametral  plane. 

Taking  the  plane  of  moments  as  y^-plane  and  choosing 
as  the  element  of  volume  a  circular  disk  parallel  to  the 
?/2-plane,  we  have 

Iy^  =  TT  \    x^y^  dx  =  TT  \    x^ijjp'  —  x'^^dx  =  t\  ira^  =  -J-  Ma?. 


EXERCISES 
Find  the  following  moments  of  inertia. 

1.  The  following  system  of  particles,  with  respect  to  each  of  the 
coordinate  planes  :  3  units  at  (0,  0,  2),  2  units  at  (4,  3,  2),  4  units  at 
(-  2,  2,  1),  1  unit  at  (3,  -  3,  0). 

2.  A  right  circular  cylinder  with  respect  to  the  plane  of  the  base. 

3.  A  paraboloid  of  revolution  bounded  by  a  plane  through  the 
focus  at  right  angles  to  the  axis,  (a)  for  that  plane ;  (h)  for  the  plane 
tangent  at  the  vertex. 

4.  A  spherical  surface  for  a  diametral  plane,  using  (a)  cartesian, 
(6)  polar  coordinates. 

5.  A  right  circular  cone  with  respect  to  the  plane  of  the  base. 

^  Ans.   -^^Mh\ 

6.  An  ellipsoid  of  revolution  with  respect  to  the  plane  through 
the  center  perpendicular  to  the  axis. 

7.  The  lateral  surface  of  a  right  circular  cone,  for  the  plane 
through  the  vertex  at  right  angles  to  the  axis. 

8.  The  volume  in  Fig.  68,  with  respect  to  each  coordinate  plane. 

9.  The  banister  cap  of  Ex.  14,  p.  163,  with  respect  to  the  plane  of 
the  base. 

10.  A  right  pyramid  with  a  square  base,  with  respect  to  the  plane 
of  the  base. 

11.  An  ellipsoid  with  respect  to  the  three  principal  planes. 

134.  General  theorems  on  moments  of  inertia.  We  pro- 
ceed to  state  certain  theorems  by  means  of  which  the 
work  of  finding  moments  of  inertia  may  in  many  cases 
be  greatly  simplified. 


196  CALCULUS 

Theorem  I :  The  moment  of  inertia  of  any  mass  with 
respect  to  a  line  is  equal  to  the  sum  of  the  moments  with  re- 
spect to  two  perpendicular  planes  through  the  line. 

For  example, 

■^x  xy     I     -^zx' 

Corollary  :  The  moment  of  inertia  of  a  plane  mass 
with  respect  to  a  line  perpendicular  to  its  plane  is  equal  to 
the  sum  of  the  moments  with  respect  to  two  lines  in  the  plane 
intersecting  at  right  angles  in  the  foot  of  the  perpendicular. 

For  example,  for  'a  mass  in  the  a::?/-plane, 

Theorem  II  :  The  moment  of  inertia  of  any  mass  with 
respect  to  a  line  (^or  plane)  is  equal  to  the  moment  with 
respect  to  the  parallel  centroidal,  line  *  (or  plane)  plus  the 
product  of  the  mass  by  the  square  of  the  distance  between  the 
lines  (^or  playies). 

That  is,  if  I  is  any  line,  T  the  parallel  centroidal  line,  d 
the  distance  between  them,  then 

Ii  =  lT-^Md\ 

We  shall  prove  these  theorems  at  present  only  for  a 
system  of  particles,  returning  to  the  general  case  later 
(§187). 

To  prove  theorem  I,  let  us  take  the  two  perpendicular 
planes  as  the  2^j/-plane  and  the  23;-plane.  Then,  for  a 
system  of  n  particles, 

n 

Ix=  ^yni{yc  +  z?~), 


Hence 


■'-xy 

2^  rriiZ?, 

n 
i=l 

I.= 

-'-xy     i~  -'-zx' 

* 

That  is, 

the 

parallel 

line  through 

the  centroid. 

CENTROIDS.     MOMENTS  OF  INERTIA 


197 


The  proof  of  the  corollary  is  left  to 
the  student. 

To  prove  theorem  II  for  any  line  I 
and  the  parallel  centroidal  line  I,  we 
note  that  in  Fig.  77,  by  the  cosine  law, 

d^^d^-^cT^-lddi  cos6>^ 

where  pi  is  the  distance  of  nii  from  the 
plane  through  I  perpendicular  to  the 
plane  of  I  and  I.     Hence 


Fig.  77 


II  II  n  n 

^  rriidi-  =  ^  viid^^  +  d^^mi  —  2d^  miPi. 

i—i  i=\  •  i—i  i=i 

n 

But  the   quantity  ^  ^^^ipt  ^^  the  mass-moment   of  first 


i=i 


order  of  the  system  with  respect  to  the  plane  through  I 
perpendicular  to  the  plane  determined  by  I  and  I,  and  since 
this  perpendicular  plane  contains  the  centroid,  the  moment 
in  question  is  0.     Hence 

1,=:  !-,-{.  Md\ 

The  proof  of  theorem  II  for  two  parallel  planes  is  still 
simpler.     It  is  left  to  the  student. 


EXERCISES 

Find  the  following  moments  of  inertia. 

1.  A  right  circular  cylinder  with  respect  to  (a)  a  plane  through 
the  axis  ;  (b)  a  generator ;  (c)  a  diameter  of  the  middle  section ; 
(<f)  a  line  tangent  to  the  base. 

2.  A  cube  with  respect  to  (a)  one  face;   (b)  an  edge. 

3.  A  circular  disk  (a)  for  a  tangent,  (b)  for  a  perpendicular 
through  a  point  in  the  circumference.     Solve  (6)  in  two  ways. 

4.  An  isosceles  triangle  about  a  line  (a)  parallel  to  the  base 
bisecting  the  altitude,  (b)  through  the  vertex  perpendicular  to  the 
plane. 

5.  A  sphere  with  respect  to  a  tangent.     Solve  in  two  ways. 


198  CALCULUS 

6.  A  square  plate  for  a  line  perpendicular  to  its  plane  (a)  through 
a  corner,  (b)  through  the  center. 

7.  A  right  pyramid  with  a  square  base,  with  respect  to  the  axis. 

8.  The  area  in  Fig.  75  with  respect  to  a  line  through  the  centroid 

(a)  parallel  to  the  base,  (b)  perpendicular  to  the  plane. 

9.  A  wire  frame  in  the  shape  of  an  isosceles  triangle,  with  respect 
to  a  line  (a)  through  the  centroid  parallel  to  the  base,  (b)  through 
the  vertex  perpendicular  to  the  plane. 

10.  An  ellipsoid  of  revolution  about  a  diameter  of  the  middle  cross 
section. 

11.  The  area  in  Fig.  76  with  respect  to  a  line  through  the  centroid 
parallel  to  the  base. 

12.  A  right  circular  cone  with  respect  to  (a)  a  diameter  of  the  base, 

(b)  a  line  through  the  vertex  perpendicular  to  the  axis,  (c)  a  diameter 
of  the  middle  cross  section. 

Ans.    (&)(|A2+,3_«2)3/.    ^c)  (j\h^-}-^\a^)M. 

13.  The  volume  in  Fig.  68,  with  respect  to  each  coordinate  axis. 

14.  An  ellipse  for  a  line  through  the  center  perpendicular  to  the 
plane. 

15.  An  ellipsoid  for  each  of  its  axes. 

135.  Kinetic  energy  of  a  rotating  body.  The  kinetic 
energy  of  a  particle  of  mass  m  moving  with  a  velocity  v  is 
defined  as 

If  a  particle  at  a  distance  r  from  a  fixed  line  rotates 
with  an  angular  velocity  &)  about  that  line  as  an  axis,  its 
linear  velocity  is  (§  58) 
V  =(Dr\ 
hence  its  kinetic  energy  is 

The  kinetic  energy  of  a  system  of  n  particles  rotating  with 
angular  velocity  w  is 

n 
1=1 

where  I  is  the  moment  of  inertia  of  the  system  with  respect 
to  the  axis  of  rotation.     This  formula  holds  in  general. 


CENTROIDS.      MOMENTS  OF  INERTIA  199 

A  discussion  of  the  various  systems  of  units  in  actual 
use  is  outside  the  scope  of  this  book.  In  the  exercises 
below  the  so-called  engineering  systeni  is  used.  In  this 
system  the  mass  m  is  replaced  by  the  value 

w 

m  =  — , 

9 
where  w  is  the  "  weight "  in  pounds,  and  g  the  acceleration 
of  gravity  (32  ft.  per  second  per  second  approximately). 
If  then  V  is  expressed  in  feet  per  second,  the  energy  is 
measured  in  ''  foot-pounds." 

EXERCISES 

1.  A  straight  rod  10  ft.  long,  weighing  20  lbs.,  rotates  about  a 
perpendicular  through  one  end  at  the  rate  of  2  R.  P.  S.  Find  its 
kinetic  energy.  Ans.    1650  ft.-lbs. 

2.  A  flywheel  12  ft.  in  diameter  whose  rim  weighs  10  tons  makes 
50  R.  P.  M.  Neglecting  the  mass  of  the  spokes,  find  the  kinetic 
energy  of  the  .wheel.  Ans.    156  ft.-tons. 

3..  A  flywheel  1  ft.  in  diameter,  weighing  50  lbs.,  makes  100  R.  P.  M. 
If  the  wheel  can  be  considered  as  a  uniform  circular  disk,  find  its 
kinetic  energy. 

4.  A  wheel  4  ft.  in  diameter  has  8  spokes  weighing  20  lbs.  each. 
The  rim  weighs  600  lbs.  Find  the  kinetic  energy  of  the  wheel  when 
it  is  making  20  R.  P.  M. 

5.  The  kinetic  energy  of  a  solid  sphere  1  ft.  in  diameter  making 
60  R.P.M.  about  an  axis  through  its  center  is  5  ft.-lbs.  Find  the 
weight  of  the  sphere.  Ans.    80  lbs. 

6.  A  hollow  cast-iron  sphere  (sp.  gr.  7.5)  4  ft.  in  diameter  rotates 
about  an  axis  through  its  center  at  the  rate  of  1  radian  per  second. 
Its  kinetic  energy  is  375  ft.-lbs.     Find  the  inner  radius.         Ans.  1  ft. 

7.  Find  the  inner  radius  in  Ex.  6  if  the  kinetic  energy  is  200 
ft.-lbs. 


CHAPTER   XIX 


LAW  OF  THE  MEAN.     EVALUATION  OF  LIMITS 


Fig.  78 


136.  Rolle's  theorem.  Let  there  be  given  a  continuous, 
one-valued,  and  differentiable  function  </)(a;),  which  van- 
ishes a.t  X  =  a  and  x  =  h.  In 
order  that  the  function,  start- 
ing with  the  value  0  at  x  =  a^ 
shall  assume  again  the  value  0 
at  rr  =  6,  it  must  first  increase 
up  to  some  point  P,  and  then 
begin  to  decrease,  or  vice  versa. 

At  P  there  is  either  a  maximum  or  a  minimum,  and  the 
derivative  is  0  at  that  point. 

We  have  thus  *  »  O 

Rolle's  Theorem  :  If  ^(a;)  -^mntskes  when  x  =  a  and 
X  =  b^  then  ^'  (x)  vani&hes  for  at  least  one  value  ofx  between 
a  and  b.  "*"  '--^ 

If  the  fundamental  assumptions  of  continuity  and  dif- 
ferentiability are  not  satisfied,  the  theorem  may  not  hold. 
In  Fig.  79  it  falls  be- 
cajj^e  <pQ^y  is  discon- 
tinuous at  one  point ; 
in  Fig.  80  it  fails  be- 
cause the  derivative 
is  undefined  at  one 
point. 

137.  The  law  of  the  mean.  Let;  f(^x}  be  a  continuous, 
one- valued,  and  differentiable  function  whose  graph  in 
the  interval  x  =  a  to  x  =  b  is  shown  in  Fig.  81.  It  is 
geometrically  obvious  that  at  some  point  P  the  tangent 

200 


Fig.  79 


Fig.  80 


LAW  OF  THE  MEAN 


201 


must  be  parallel  to  the  secant 
SQ.  Now  the  slope  of  the 
secant  is 

SR  h-a       ' 

the  slope  of  the  tangent  at  P 
is  /'(^i),  where  x^  is  the  ab- 
scissa of  P.     Hence   ^^^o^""*^^' 

or 

(1)  iW-Ka)  =  (h-a)}'ixi).         a<x,<b. 

This  relation  is  called  the  law  of  the  mean. 

138.  Other  forms  of  the  law  of  the  mean.  It  is  often 
necessary  to  apply  formula  (1)  above  with  b  =  x,  thus 
making  the  length  of  the  interval  variable  : 

( ■'-)  £i^)  =  •^(^)  +  ^^  ""  ^)/'C^i)'         a<x^<x. 

It  is  to  be  noted  that  x-^^  is  here  a  function  of  x. 
Again,  placing 

a  =  x^     h  —  a  -\-  Ax^ 
and  denoting  by  ^  a  positive  number  less  than  unity,  we 
obtain 

f(x  4-  Ax)  =  f{x)  +  Axf'(x  +  OAx).         0  <  ^  <  1. 

EXERCISES 

For  each  of  the  following  functions,  show  why  Rolle's  theorem 
does  not  hold  in  the  indicated  interval. 


1.  y 


-^- 


-i<.r<i. 


2.    (y  +  4y=x%  -8<a;<8. 

^    y  =  tan  x,  0  <  x  <  tt. 

4.  Draw  curves  showing  that  the  law  of  the  mean  may  fail  when 
f(x)  is  discontinuous  or  non-differentiable  in  the  interval. 

5.  At  what  point  on  the  parabola  y  =  x^  is  the  tangent  parallel  to 
the  secant  through  the  points  x  =  0,  x  =  2  ? 


202  CALCULUS 

6.  At  what  point  on  the  curve  y  =  x^  -  x\s  the  tangent  parallel  to 
the  secant  through  (1,  0)  and  (2,  6)?     Draw  the  figure. 

7.  Fmd  the  point  on  the  curve  y  =  log  x  where  the  tangent  is  paral- 
lel to  the  secant  through  the  points  x  =  1,  a;  =  2. 

139.    The  indeterminate  forms -,  g-     If   two   functions 
/(a;),  F(x)  both  vanish  Sit  x  =  a: 

/(a)=0,  l^(a)  =  0, 

their  quotient  -J-  ^    assumes    the    "  indeterminate   form  " 
^  Fix) 

-  Sit  X  =  a,  and  is  undefined  at  that  point.     Nevertheless 
the  limit  lim  -^  ^       may  exist.      This  fact  is  illustrated  in 

.^a  F(X)  ^ 

the  derivation  of  the  fundamental  differentiation  formulas, 
where  in  each  case  both  numerator  and  denominator  of 

the  difference  quotient  —^  approach  0,  yet  the  derivative, 

which  is  the  limit  of  that  quotient,  exists. 

f  (x^ 
In  case  the  fraction  ^„^  ^  does  approach  a  limit  when 

Fix)  ^^ 

X  approaches  a,  we  lay  down  the  following  definition: 


/M 

Fix)] 


=  lim  ^  ^  ^ 

x=a        x-^a  F  {x^ 


The  function  ^  ^  ^  thus  becomes  continuous  at  the  point 
F(x')  ^ 

x  —  a^  by  the  definition  of  continuity  (§  12). 

It  may  be  possible  to  evaluate  the  limit  by  means  of 

more  or  less  obvious  transformations  of  "^^  ■  ,  as  was  done 

FQx^ 

in  deriving  the  differentiation  formulas.  In  many  cases 
the  limit  may  be  obtained  by  a  method  that  will  now  be 
developed. 

By  the  law  of  the  mean,  using  the  form  (1)  of  §  138 


I 


LAW  OF  THE  MEAN  203 

we  may  write 

F(x)  =  F{a)  +  (a:  -  a)F'(x,^), 

where  x^  and  X2  lie  between  a  and  x.     But  by  hypothesis 

/(a)  =  l^(a)=0. 

Hence 

f(:x}^(x-a}f'(x,)^f'Cx,} 

.       F{x)      (x-a)F'(x2^      F'ix^) 

As  X  approaches  a,  x^  and  x^  must  do   likewise,  and  we 
have,  by  theorem  III  of  §  8, 

lim  -^  ^    ^  =  lim '- — ^^-1^  =  lim  '^  /   -    =  ■^  ,^   ^  , 

x^aF(x)         x^aF'ix^)         x-^a  F' (x)         F\a) 

provided  f'(a)  and  F'(^a)  exist  and  F'{a')^0. 

If /(a-)  and  J^ (a;)  both   increase   indefinitely   as  x  ap- 
proaches a : 

lim/(2;)  =  QO,  lim  JP(a;)=  Qo, 

the  fraction  "i,^    ^  is  said  to  assume  the  indeterminate  form 
Fix) 

^  3.t  X  =  a.     Here  again  it  may  happen  that  lim  \}      ex- 

'^  x-^a  Fix) 

ists,  and  it  can  be  shown  that  the  same  method  may  be 

applied  in  this  case  as  in  the  case  just  treated. 

f  (x) 
It  may  happen  that  the  fraction    '^^^      takes  the  form 

^       ^^  F'(x) 

-  or  ^.     In  this  case  we  may  differentiate  numerator  and 

denominator  again,  and  repeat  as  many  times  as  necessary. 

Finally,  we  may  have  a  —  co  :  i.e.  ~y^  approaches  the 

form  -  or  ^  when  x  increases  indefinitely.     In  this  case 

the  rule  holds  also. 

The  results  of  this  article  may  be  summarized  in  the 
following 


204  CALCULUS 

fix] 
Fix) 


f(x') 
Theorem  :     If  the  fraction  ^  ^  ^  assumes  the  indetermi- 


nate form  -  or  —  when  x  —  a.  then 

•^  Q  OO  ' 

lim  '^  ^   ^  z=  lim  •^    ^  ^  , 

x^a  F (X)         x-^a  F'{X) 

provided  the  latter  limit  exists. 

Thus  we  may  differentiate  the  numerator  and  the  de- 
nominator separately^  and  take  the  limit  of  the  new 
fraction  thus  formed.  It  must  be  borne  clearly  in  mind, 
however,  that  the  theorem  applies  only  to  fractions  in 
which  the  numerator  and  the  denominator  hoth  approach 
0  or  hoth  increase  indefinitely. 

Example:  Evaluate  lim 

^->o      X 

This  fraction  takes  the  form  -  when  ri;  =  0,  so  that  the 

theorem  applies : 

1 .      tan  X      1 .      sec^  x      ^ 
lim =  lim  — — -  =  1. 

a:->0        X  x-^0        1 

140.  The  indeterminate  forms  0  •  go  ,  oo  —  oo.  Given 
the  product  of  two  functions /(a;)  •  FQx),  suppose  that  as 
X  approaches  a  one  function  approaches  0  while  the  other 
increases  indefinitely.  The  product  is  then  said  to  take 
the  indeterminate  form  0  •  oo . 

If  we  write 

fix)  .  Fix)  =  ^, 

Tix) 
it  appears   that   the   quotient   last   written   assumes  the 

form       or  ^,  and  the  theorem  of  §  139  may  be  applied. 

If,  as  X  approaches  a,  each  of  two  functions /(a;),  F(x) 
increases  indefinitely,  their  difference  fXx')—  F(x)  is  said 
to   assume   the  indeterminate   form   oo— oo.      Here    also 


LAW  OF  THE  MEAN  205 

we  express /(^)  — ^(2:)  as   a  fraction    which   takes    the 
form  -  or  ^ ,  and  then  apply  the  theorem. 

Example  :  Evaluate  lim  x  log  x. 

This  takes  the  form  0  •  go.     If  we  write  it  in  the  form 


— 5^,  the  theorem  of  §  139  becomes  applicable : 

~x                                                          1 
lim  X  log  X  =  lim  — ^ —  =  lim  =  lim  (  —  2:)  =  0. 


X  X^ 

EXERCISES 

Evahiate  the  following  limits,  when  they  exist. 


1. 

1.  .     cos  X 
lini  • 

^   TT                     IT 

2^      2 

3. 

lim 

x->4  x'^  +  X  —  20 

5. 

lim   sin 2  x 

a— >0 

X 

7. 

lim    a;  log  sin  a;. 

9. 

lim       3x^-4^    , 
x^^  2^2-  3a;  +  1 

11. 

lim  ^'-4:^3 

13. 

lim    ^  —  arcsin  ^ 
e^-o        sin^  6 

Trace  the  following  curves. 

14. 

y  =  X  log  a:. 

16. 

lOQ'  X 

V  =  ■ — 

2. 

lim  —  • 

4. 

x->-i      a:  +  1 

6. 

lim  a:e~*. 

a— ^cc 

8. 

lim   log  cos  a: 
x->o         a; 

10. 

T        a:  —  1 

lim 

x->x  a:-'^  +  1 

12. 

lim  (sec  x—  tan  x). 

^n5.      --. 
6 

15. 

?/  =  are~*. 

17. 

y  =  X'  log  x. 

18.  Find  the  area  in  the  fourth  quadrant  bounded  by  the  curve 
y  =  log  X  and  the  coordinate  axes. 

19.  Find  the  centroid  of  the  area  in  the  second  quadrant  under  the 
curve  y  —  e^.     Obtain  each  coordinate  in  two  ways. 


206  CALCULUS 

20.  Find  the  moment  of  inertia  of  the  area  in  Ex.  19,  about  each 
coordinate  axis. 

21.  Find  the  area  bounded  by  the  curve  y  =  xlog  x  and  the  x-axis. 

141.  General  remarks  on  evaluation  of  limits.  While 
the  methods  of  §§  189-140  are  frequently  very  useful  in 
investigating  the  limit  of  a  function  at  a  point  where  the 
function  ceases  to  be  defined,  they  are  by  no  means 
always  applicable.  In  the  first  place,  the  function  may 
lose  its  meaning  in  some  other  way  than  by  taking  the 

form  TT   or   ^   (or  a  form  reducible  to  one  of  these),  so 

that  the  theorem  of  §  139  cannot  be  brought  into  play, 
yet  it  may  be  possible  to  show  the  existence  of  a  limit  by 

other  methods.     Even  when  the  function  '^j^   ^   does  take 

0  ^^""^ 

the  form   -    or   ^  the  theorem  may  fail  to  apply  because 

'^—^ — -  approaches  no  limit,  yet  the  limit  of  the  original 

quotient  may  exist.     Again,  the  function  ^_^  ^  may  take 

0  /'  (x) 

the  form  i^   ov  —^  and  at  the  same  time   -^,,  ;  may  ap- 

proach  a  limit,  yet  it   may  be  impossible  to  obtain  any 

result   by   the   use    of    the    theorem.      Finally,    there   is 

always  the  possibility  that  a  function  undefined  at  a;  =  a 

may  fail  to  approach  any  limit  as  x  approaches  a. 

Each  of  these  cases  is  illustrated  by  the  following 

sin  X 
Examples:  (a)  Evaluate  lim • 

Here  the  denominator  increases  indefinitely,  while  the 

numerator   oscillates   between    —  1    and    1,    without    ap- 

sm  X   ' 
proaching  any  fixed  value.     Nevertheless,  since  is 

X 

never  numerically  greater  than  -,  it  clearly  approaches  0. 

37 


LAW  OF  THE  MEAN  207 

1 


a:^  cos  - 

(b)  Evaluate  lim — : 

x-^o    Sin  X 

Since  cos  -  lies  always  between  —  1  and  1,  the  numer- 

0 
ator   approaches  0    and    the   fraction  takes  the   form   -x  • 

Differentiating    numerator   and    denominator   separately, 

z  X  cos  -  +  sm  - 

X  X 

we  obtain  a  new  fraction Tliis  fraction 

cos  X 

approaches   no  limit,  since  sin  -   oscillates  between    —  1 

and  1,  and  the  theorem  is  therefore  inapplicable.  But 
the  original  fraction  has  a  limit,  which  may  be  found 
directly : 

X'^  COS  - 

X  .  X  '  1 

lim  — : =  lim  — •  lim  x  cos  -  =  0, 

x^o    ^^^  ^         -r->o  sin  X     x-^o  X 

X  1 

since   lim  -^—  =  1  and  lim  x  cos  -  =  0. 

x->o  sm  X  x->o  X 

Ox 

(c)  Evaluate  lim  —  • 

X-^co  O 

This  fraction  takes  the  form  —-     By  §  139, 

y      2^      ,.      2- log  2      T      2^1og2  2 

lim  —  =  lim  & — =  lim  ^ — =  ••-. 

a>->oo  3^      a>->oo  3^  log  3      ^-».cc  3^  log^  3 


No  matter  how  many  times  we  differentiate,  we  cannot 

2^ 
get  rid  of  the  quotient  — •     Yet  if  the  function  be  writ- 
ten in  the  form  (f)^  it  is  seen  to  approach  the  limit  0. 
(c?)  Evaluate   lim 

x-^co       X 

In   this  case  no   limit  is  approached.     For  no  matter 
how  large  x  be  taken,  as  x  varies  from  /^tt  to  (n  -|-  l)7r 


208  CALCULUS 

the  function  tan  x^  and  hence   ,  ranges   through  all 

possible  values  from  —  go  to  +20. 

EXERCISES 

a. 

Evaluate  the  following  limits,  when  they  exist. 
1-    1^"^^/  ^ris.   0.  2.     lim  sin  <^ cot  <^.     Ans.    -1. 


n 

M->00  2' 

cos  X 


x-^-o        X 

5. 

lim  ^°S<^1  +  ^') 

a— ^00                .y 

7. 

,.      tan  X  —  X 
lim  ^ 

a:_>0  ^*  —  Sni  X 

9. 

lim     '^^^^   . 

j;_>oo  cos  2  X 

1 

^•,^_^  sin  a;  —  sin  a 

^ns.    0.  4.     lim  (a:^  -  x). 

6.     lim  ' 

8.    lim  X  sin  -  • 

x->0  X 

tan  ^ 


10.     Hm 


«_.  flrtan  3  B 
12.    Imi 


x_^a         X-  a  x->i  a;3  -  a-2  _  a;  +  1 

1 

a;  cos  - 

13.    lim  — : 14.    lim  (cot  a:  —  cosec  a:) . 

x-^Q    sm  a;  a-^-o 

15.     lim^/?^±i.  16.    lim  —.  Ans.   0. 

Trace  the  following  curves. 

^„            sin  a;                                         ^o            tana; 
17.   y  = 18.   y  = 

X  "  X 

19.   y  =  x^e-''.  20.   y  =—- 

X 

lo£r  a; 

21.  Find  the  area  bounded  by  the  curve  y  =  ,  the  a:-axis,  and 

the  maximum  ordinate.     Trace  the  curve.  Aris.   |. 

22.  Trace  the  curve  y  =  — ^ ,  and  find  the  area  under  the  curve 

X  log  X 

from  X  =  2  to  a:  =  e.  Ans.   0.367. 

23.  Trace  the  curve  y  =  xe   ^    ,  and  find  the  area  under  the  curve 
in  the  first  quadrant. 

24.  Find  the  moment  of  inertia  of  the  area  in  Ex.  23,  with  respect 
to  each  coordinate  axis. 


CHAPTER   XX 

INFINITE  SERIES.    TAYLOR'S  THEOREM 
I.    Series  of  Constant  Terms 

142.  Series  of  n  terms.  A  series  of  n  terras  is  an  expres- 
sion of  the  form 

^1  +  ^2  +  ^3+ 1"  ^»' 

where  each   term  is   formed  from  the  preceding  one   by 
some  definite  law.     Examples  are  the  arithmetic  series 

a  -{-[a  -h  d]  -\-  [a  -h  2  d~\  +  •••  -{-[a  +  ^n-  l)d], 
in  which  each  term  is  formed  from  the  preceding  by  the 
addition  of  a  constant  cZ,  and  the  geometric  series 

a  -{-  ar  -\-  ar^  +  •  •  •  +  ar""^, 
in  which  each  term  is  equal  to  r  times  the  one  before  it. 

143.  Infinite  series.  When  the  number  of  terms  in- 
creases indefinitely,  a  series  of  n  terms  becomes  an  infinite 
series^  denoted  by  the  symbol 

aj  -f  «2  +  ^3  +  •••• 
The  series  is  defined  by  the  law  of  formation  of  successive 
terms,  or,  what  amounts  to  the  same  thing,  by  the  n-th  or 
general  term. 

The  .general  term  may  frequently  be  written  down  by 
inspection  of  the  first  few  terms,  as  in  the  following 

Examples :  (a)   In  the  series 

the  general  term  is  -• 

n 

(J)  In  the  geometric  series 

l_l-}-l_i-i-  ... 

2     "^  4         8     '  ' 

the  general  term  is  (—  J)"~^. 
p  "        209 


210  CALCULUS 

(<?)  In  the  series 

1     1      _J_     1  1  1  1  1 

2"32-4'52.4.6*72.4.  6  •  8*9      '"' 

,    the  general  term  is  ^  •  ^. 

144.    Sum  of  an  infinite  series.     It  is  shown  in  elemen- 
tary algebra  that  the  sum  of  a  geometric  series  of  n  terms  is 

(1)  s,    ''-''''^ 


1  —  r 

of  an  arithmetic  series  of  n  terms, 

where  I  is  the  last  term — i.e. 

>^„  =  ^[2a+(^-l)(^]. 

Similarly  the  sum  of  any  series  of  a  finite  number  of  terms 
can  be  found. 

On  the  other  hand,  an  infinite  series  has  no  sum  in  the 
ordinary  sense  of  the  term,  since  no  matter  how  many 
terms  we  might  add  up,  there  would  always  be  an  infinite 
number  left  over.  We  may,  however,  give  a  meaning  to 
the  term  "  sum  "  even  in  this  case  by  laying  down  the  fol- 
lowing definition  : 

The  sum  of  an  infinite  series  is  defined  as  the  limits  as  n 
increases  iiidefinitely.,  of  the  sum  of  the  first  n  terms  : 

provided  the  limit  exists. 

Thus  the  "  sum  "  of  an  infinite  series  is  the  limit  of  an 
ordinary  sum. 

Example :  By  (1),  the  sum  of  the  first  n  terms  of  the 
infinite  geometric  series 

a  -\-  ar  +  ar'^  +  •  •  •  +  a?'"~^  +  •  •  • 


is 


aSL  = 


a  —  ar 
1-r 


INFINITE  SERIES.     TAYLOR'S  THEOREM       211 

Hence  the  sum  of  the  series,  if  the  sum  exists,  is 

^__  lim  a—_ar^ 

When  r  is  numerically  less  than  1,  the  quantity  ar"  ap- 
proaches 0  as  72  increases,  and 

1  —  r 

When  r  is  numerically  greater  than  1,  the  quantity  ar"  in- 
creases indefinitely,  and  the  above  limit  does  not  exist; 
the  series  has  no  sum. 

145.  Convergence  and  divergence.  If  the  series  has  a 
sum  aS',  i.e.  if  S^  approaches  a  limit  when  n  increases,  the 
series  is  said  to  be  convergent.,  or  to  converge  to  the  value  S; 
if  the  limit  does  not  exist,  the  series  is  divergent. 

It  follows  from  the    above    example    that  a  geometric 

series  converges  to  the  value  if  |r|<  1;  it  diverges 

1  —  7' 

if  |r|  >1. 

A  series  may  diverge,  as  in  the  case  of  a  geometric 
series  for  which  r  >  1,  because  aS'„  increases  indefinitely  as 
n  increases ;  or  it  may  diverge  because  S^  increases  and 
decreases  alternately,  or  oscillates,  without  approaching 
any  limit.     In  the  latter  case  the  series  is  called  oscillatory. 

EXERCISES 

1.  Show  that  every  infinite  arithmetic  series  is  divergent. 

2.  Find  the  sum  of  a  geometric  series  of  n  terms  for  which  r  =  1, 
(a)  directly,  (6)  by  applying  the  theorem  of  §  189  to  formula  (1), 
§  144. 

3.  Show  that  the  infinite  geometric  series  for  which  r  =  1  is 
divergent. 

4.  Show  that  the  infinite  geometric  series  for  which  r  =  —  1,  viz. 

a  —  a-\-a— a-{-  •••, 
is  oscillatory. 

146.  Tests  for  convergence.  In  the  elementary  applica- 
tions divergent  series  are  of  no  importance.     Before  being 


212  CALCULUS 

» 

able  to  use  a  given  series  we  must  determine  whether  it 
converges  or  diverges.  If  S^  can  be  expressed  explicitly 
as  a  function  of  n,  as  in  the  case  of  the  arithmetic  and 
geometric  series,  we  can  in  general  determine  the  con- 
vergence or  divergence  of  the  series  directly,  and  find  the 
sum  if  it  exists ;  but  S^  cannot  be  so  expressed  in  most 
cases. 

A  necessary  condition  for  convergence  is  that  the  gen- 
eral term  approach  0  as  its  limit  : 

lim  a^  =  0. 

n— >>oo 

For,  when  this  condition  is  not  satisfied,  each  term  that  is 
added  changes  S^  by  an  amount  that  does  not  approach  0, 
so  that  the  difference  between  S^  and  a  fixed  number  S 
obviously  cannot  become  and  remain  arbitrarily  small. 

This  condition,  though  necessary,  is  not  sufficient;  i.e. 
if  the  condition  is  not  satisfied,  the  series  diverges,  but  if 
it  is  satisfied,  the  series  still  may  diverge.  This  is  illus- 
trated by  the  harmonic  series 

14-14-14-14-  ... 

which  will  be  shown  in  the  next  article  to  be  divergent, 
although 

lim  a„  =  lim  -  =  0. 

Many  special  tests  for  convergence  have  been  devised, 
applicable  to  more  or  less  broad  classes  of  series.  Several 
of  the  simplest  are  considered  in  the  next  few  articles. 

147.    Cauchy's  integral   test.     We  will  begin  with   an 
JExample:  Prove  that  the  harmonic  series 

(1)  i+i  +  j+i+- 

is  divergent. 

Here  the  general  term  is 

n 


INFINITE  SERIES.     TAYLOR'S  THEOREM       213 


Let  us  draw  the  curve 
1 


y=fi^~)  = 


X 


erect  the  ordinates  at  2:  =  1, 
2,  3,  ••.,  71,  and  complete  the 
circumscribed  rectangles  as 
shown  in  the  figure.  Then 
the  areas  of  the  rectangles 
are,   respectively, 

1 1  1 

'  2'  3' 


Fig.  82 


1 

n 


so  that  the  sum  of  these  areas  is  the  sum  of  the  first  n 
terms  of  the  series  (1)  : 

A      6  n 

But  the  sum  of  the  rectangles  is  clearly  greater  than  the 
area  under  the  curve  from  a;  =  1  to  x=n'. 


y  dx=  \     —  =  log  92. 

1  *^l      X 


When  n  increases,  the  area  log  n  under  the  curve  becomes 
infinite,  hence  S^  does  likewise  and  the  series  diverges. 

This  example  illustrates  CaucTiyB  integral  test : 

Given  a  series  of  positive  terms 

(2)  ai  +  «2  +  ^3+  *••» 

P^t  an  =  fin). 

If  the  function  f(x)  is  defined  not  only  for  positive  integral 
values,  but  for  all  positive  values  of  x,  and  if  fix)  never  in- 
creases with  X,  then  the  series  (2)  converges  or  diverges  ac- 

fQx)  dx  does  or  does  not  exist. 

The  proof  of  this  test  is  easily  written  out  by  drawing 
the  curve  y  =fQc)  and  following  the  process  suggested  by 
the  above  example.     The  details  are  left  to  the  student. 

*  For  the  definition  of  this  improper  integral,  see  §  119. 


214  CALCULUS 

EXERCISES 

1.  Write  out  the  proof  of  Cauchy's  integral  test. 

2.  In  the  statement  of  the  integral  test,  why  is  it  assumed  that 
/(x)  never  increases  with  xl  Show  that  it  would  be  sufficient  to 
assume  that/(x)  never  increases  with  x  after  some  fixed  point  x  —  x^. 

3.  Prove  that  the  series 


1.2     2.3     3-4     4.5 
is  convergent. 

4.   Prove  that  the  series 

2P      3^      4^ 

converges  if  7?  >  1,  diverges  if  p  ^  1. 

Test  the  following  series  for  convergence. 

1.2      3.4      5.6 

6.  1  +  — i — +      -*-       +      -^      +  '•'. 

1+22      1+32      1+4^ 

7.  i_2+3-4+  .-.. 

8.  1-f +i-f+  •••• 

9.  1+i  +  i  +  f +-. 

2  3  4 

10,  1  +  -A —  +  _2_  +      ^      +  .... 

^  1  +  22     1  +  32      1  +  42 

11.  Test  the  geometric  series  for  convergence  by  the  integral  test. 

148.    Comparison  test.     Let 

w  J  -h  ^2  +  ^3  +  *  *  • 

be  a  series  of  positive  terms  to  be  tested. 

(a)  If  a  series 

^  ^    ^  a^-\-  a^-\-  a^-\-  ... 

of  positive  terms,  known  to  be  convergent,  can  he  found  such 
that  ^ 

then  the  series  to  be  tested  is  convergent. 
(5)  If  a  series         1,1,7, 


INFINITE  SERIES.     TAYLOR'S  THEOREM        215 

of  positive  terms^  known  to  be  divergent^  can  be  found  such 
that  -^  1 

Un  ^  bn-> 

then  the  series  to  be  tested  is  divergent. 

To  prove  (a),  let  S^Cu)  be  the  sum  of  the  first  n  terms 
of  the  2^-series,  Sn^a}  the  sum  of  the  first  n  terms  of  the 
a-series,  and  S(^a~)  the  sum  of  the  a-series.  Since  all  the 
terms  w„  are  positive,  S^Cu')  always  increases  with  n.  On 
the  other  hand,  we  have 

S^Cu')  <  S^a')  <  S(a'). 

Since  S^^u)  always  increases,  but  never  exceeds  the  fixed 
number  >S'(a),  it  approaches  a  limit,  by  theorem  IV  of  §  8, 
which  is  not  greater  than  SQa'). 

The  proof  of  (b)  is  left  to  the  student. 

The  success  of  the  test  depends  on  our  ability  to  find 
a  convergent  series  whose  terms  are  greater  than  the 
corresponding  terms  of  the  series  to  be  tested,  or  a  diver- 
gent series  whose  terms  are  less  than  those  of  the  series  to 
be  tested.  To  show  that  the  terms  of  the  w-series  are 
greater  than  those  of  some  convergent  series,  or  less  than 
those  of  some  divergent  series,  proves  nothing. 

It  is  clear  that  the  convergence  of  a  series  is  not 
affected  by  discarding  any  finite  number  of  terms  from 
the  series.  Hence  the  conditions  of  the  test  do  not  need 
to  be  satisfied  from  the  very  beginning  of  the  series,  but 
only  after  a  certain  pointy  all  the  terms  up  to  that  point 
being  neglected. 

Example  :  Test  the  series 


The  series 


^1.2     2.3     3.4 


'22      32     42 


is  known  to  converge  (Ex.  4,  p.   214).     Discarding  the 


216 


CALCULUS 


first  term  of  the  series  to  be  tested,  we  have 

1 


u„  = 


n(n  -h  1) 

The  general  term  of  the  known  series  is 

1 


a„  = 


n^ 


Since 

the  series  in  question  converges. 


EXERCISES 

Test  the  following  series  as  to  convergence  or  divergence. 

2. 


1.   1+  — +  — + 

2 !      3 ! 


1     4-A  +  A  + 


1,1.1, 

v'2      V3      V4 

K      1    ,        1        ,        1        ,        1        , 

2      2-22      3-23      4-2* 


1-2      34      5-6 

4.   1  +  1  +  U.... 
3     5     7 

6.    1  +1  +  1+.... 
3-^      52 


149.    Ratio  test.     There  are   many  ''ratio   tests";    the 
simplest  is  the  following  : 
Given  the  series 

to  be  tested   for  convergence,  form   the  ratio  — ^^  of   a 
general  term  *  to  the  one  preceding  it. 


U, 


(a)  If  lim 


(J)    If 


lim 


<  1,  the  series  converges. 


u. 


the  series  diverges. 


>1,  or  if 


u 


n+l 


Ur 


increases  indefinitely/. 


(«)  If  1'™ 


}J_^00 


u 


n+l 


Ur 


=  1,  the  test  fails. 


*  We  may  divide  the  (n  +  l)-th  term  by  the  w-th,  the  (71  +  10)-th  by 
the  (n  +  9)-th  —  a7iy  general  term  by  the  one  before  it,  since  the  question 
of  convergence  is  not  affected  by  dropping  any  finite  number  of  terms. 


INFINITE  SERIES.     TAYLOR'S  THEOREM        217 

This  test  holds  for  any  series  whatever,  not  merely  for 
series  of  positive  terras. 


Suppose   we   have   case   (aj :     ^^^'^ 


u 


n+l 


=  L<1.      At 


present  we  shall  consider  only  the  case  in  which  all  the 
terms  are  positive,  and  show  later  ' 

how  the  proof  may  be  completed.      r-- 4- 1 — ;- 

Let  us  choose  some  number  r  be-  -p^     go 

tween  L  and  1.     By  the  definition 

of  limit,  the  difference  between  the  ratio  _!^  and  its  limit 

L  ultimately  becomes  and  remains  as  small  as  we  please ; 
therefore  a  number  m  can  be  found  such  that  for  all  values 
oi  n^m  we  have 

'^n-f-l     ^ 

<  r. 

Hence 

'^m+3  <C  '^m+2^'  "^  '^mV  •> 

•  •  • 

Discarding  the  first  m  terms  of  our  series,  we  see  that  the 
remaining  terms  are  less  than  the  corresponding  terms  of 
the  series 

u^r -{- u^r^  +  u,y -\ . 

But  this  latter  series,  being  a  geometric  series  with  ratio 
r  <  1,  is  convergent  ;  hence  the  given  series  is  conver- 
gent, by  the  comparison  test. 

Case  (5)  may  be  proved  by  showing  that  the  general 
term  w„  does  not  approach  0  when  n  increases  indefinitely. 
The  details  are  left  to  the  student. 

The  test  may  be  shown  to  fail  in  case  (c)  by  the  follow- 
ing example  :   For  the  series 

2p      3^ 


218 


CALCULUS    • 


the  test  ratio  is 


T 


n 


_n  4- 


tT= 


1+1 


and 


11  111 

1  ■ 

% >'0O 

1+1 

L       ^J 

=  1. 


But,  by  Ex.  4,  p.  214,  when  jt?  <  1,  this  series  diverges  ; 
when  jo  >  1,  the  series  converges.  Hence  there  are  both 
convergent  and  divergent  series  for  which  the  limit  of  the 
test  ratio  is  1. 

150.  Alternating  series.  A  series  whose  terms  are 
alternately  positive  and  negative  is  called  an  alternating 
series.  Such  series  are  of  frequent  occurrence.  Their 
most  important  properties  are  contained  in  the  following 
theorems. 

Theorem  I  :  If  after  a  certain  point  the  terms  of  an 
alternating  series  never  increase  numerically^  and  if  the 
limit  of  the  n-th  term  is  0,  the  series  is  convergent. 

Theorem  II :  In  a  convergent  alternating  series^  the 
difference  between  the  sum  of  the  series  and  the  sum  of 
the  first  n  terms  is  not  greater  numerically  than  the  (n-\-V)-th 
term : 

l^~^n|  ^  l^n+ll- 

To  prove  theorem  I,  let  us  write  the  series  in  the  form 

Wj  — 1^2  +  ^3  —  ^4+  rt-^ 
where   all   the   u'^   are   positive.     When  n  is  even,   say 
II  =  2  m,  we  may  write  S^  in  the  two  forms 

(1)  ^2m  =  (^1  -  ^2)  +  (i^3  -  W4)  +    •  •  •    +  (^2m-l  -  '^2m)'> 

(2)  Sor^i  =  '^^1  —  (^2  —  Wg)  —    •  • .   —  (U2,a~2  "  '^2m-l)  "  %m- 

Equation  (1)  shows  that  82^  always  increases,  since  each 


INFINITE  SERIES.     TAYLOR'S  THEOREM        219 

of  the  parentheses  is  positive.  Equation  (2)  shows  that 
S2jn  is  always  less  than  u^  Since  ^2ff»  always  increases, 
but  never  exceeds  the  fixed  number  u^^  it  approaches  a 
limit  S  not  greater  than  i^j,  by  theorem  IV  of  §  8. 
Further,  the  sum  of  an  odd  number  of  terms  S2m+i  ^-p- 
proaches  the  same  limit,  since 

lim  (S.,^+^-  jS^,,,)  =  lim  W2m+i  =  0- 

Theorem  II  follows  at  once.  For,  if  n  is  even,  the  dif- 
ference S  —  Sn  is  the  alternating  series 

and  we  have  just  shown  that  the  sum  of  an  alternating 
series  is  not  greater  than  the  first  term.     Similarly  if  n 
-^is  odd. 

151.  Absolute  convergence.  A  series  is  said  to  be  abso- 
lutely  coyivergent  if  the  series  formed  from  it  by  replacing 
all  its  terms  by  their  absolute  values  is  convergent.  It 
can  be  shown  that  a  series  always  converges  if  the  series 
of  absolute  values  converges.  From  this  fact  the  proof 
in  §  149  is  easily  completed  for  the  case  when  the  terms 
are  not  all  positive. 

EXERCISES 

Determine  whether  the  following  series  are  convergent  or  divergent. 


1. 

2      2^     2^ 

3. 

fi'l         /p3        ^4 

-   +  -  +  -+    —. 

2       3      4 

5. 

1 !       2 !       3! 

10         102         103 

7. 

l-i   +  i-i+- 

9. 

l_l+l_i+ 
92        93         94 

2. 

l-l+i--.... 
2!      3! 

4. 

4        V4/         V4/ 

6. 

1      1.3      1.3.5 
3      3-6      3.6.9 

8. 

1_1+1_1+.... 

32  ^52     72  ^ 

LO. 

l-i  +  i-f  +  .". 

11.  Are  the  series  in  Exs.  7-10  absolutely  convergent  ? 

12.  Carry  out  the  proof  of  case  (h),  §  149. 


220 


CALCULUS 


y\\.    Power  Series 

152.  Power  series.  Up  to  this  point  we  have  con- 
sidered only  series  whose  terms  are  constants.  The  case 
of  greatest  practical  importance,  however,  is  that  in  which 
the  terms  are  functions  of  a  variable.  In  what  follows, 
we  shall  be  chiefly  concerned  with  the  class  known  as 
power  series. 

A  series  of  the  form 

a^-\-  a-^  ■\-  a^x^  -{-  •••, 

where  a;  is  a  variable  and  a^,  a^,  a^^  •••  are  constants,  is 
called  a  power  series.  Such  series  are  of  especial  impor- 
tance in  practice. 

A  power  series  may  converge  for  all  values  of  the  variable 
x^  or  for  no  values  except  0 ;  but  usually  it  will  converge  for 
all  values  in  some  finite  interval,  and  diverge  for  all  values 
outside  that  interval.  The  interval  of  convergence  always 
extends  equal  distances  on  each  side  of  the  point  2;  =  0. 

The  interval  of  convergence  can  usually  be  determined 
by  the  ratio  test.     We  illustrate  the  process  by  an 

Example :  Find  the  interval  of  convergence  of  the  series 


Here 


1  +  0;  +  ^+-  + 
2       3 


x^ 
n 

n+l 


X 


+  —  + 


n 


lim 


u 


n+l 


Ur 


=  lim 


X 


71  +  1 


X"^ 


n 


=  lim  - 

n-^ao  n-j-  1 


x\  =  \x\ 


by  the  theorem*  of  §  139. 

*  Objection  may  be  raised  to  the  use  of  this  theorem  in  the  present 
instance,  on  the  ground  that  n  is  not  a  continuous  variable.  The  objec- 
tion, liowever,  is  easily  disposed  of.  For,  if  we  can  prove  by  the  theorem 
that  a  given  function  of  n  approaches  a  certain  limit  when  n  varies  con- 
tinuously, it  is  certain  that  the  same  limit  will  be  approached  when  n 
varies  through  positive  integral  values. 


INFINITE  SERIES.     TAYLOR'S  THEOREM        221 

(a)  The  series  converges  when  \x\  <  1,  i.e.  —  1<  x  <,  1. 
(5)   The  series  diverges  when  \x\  >  1. 
(c)   The  test  fails  when  x=  ±1.     But  when  x  =  l  the 
series  is  the  harmonic  series 

and  therefore  diverges ;   when  x  =  —  1^  the  series  is 

which  converges  by  §  150. 

Hence  the  interval  of  convergence  is  —  1  <  a;  <  1. 

EXERCISES 

Find  the  interval  of  convergence  of  the  following  series. 

1.  1  +  X  +  x^  i-  x^  +   •••. 

2.  1-  2a;  -h'dx-^  -  i  x^  -\-    -. 

3.  i_?  +  ^_^+  .... 

3       9      27 

4.  .r  —  —  H •••.  ^n5.     All  values  of  x. 

3 !      5 ! 

5.  1  +  lOar  +  2  .  .100a:2  +  3  •  1000 x^  +  •••. 

6.  1  +  a:  +  2!a;2+  3!x3+  .... 

7.  1  +  nx  +  ^-(iLlll)  x2  4-  K^-l)(>^-2)^3  +  .... 

2!  3! 

2  3 

8.  1 +a: +  — +  — +  ••••  ^ns.     All  values  of  a:. 

2 !      3 ! 

9.  If  a^  +  Og  +  03  +  •••  is  an  absolutely  convergent  series  and 
&p  &2»  ^jp'''^;  set  of  numbers  that  remain  finite  as  n  increases: 
I  ft„  I  <  M,  where  M  is  a  constant,  show  that  the  series 

a  A  +  «2^2  +  «3^3  +  ••• 
converges  absolutely. 

10.  Prove  that  the  series 

sin  3  X  ,  sin  5  x 

sm  X 1 •  —  .  •  • 

32  5-^ 

converges  absolutely  for  all  values  of  x. 

11.  If  a^  +  flo  +  (Is  +  •••  is  an  absolutely  convergent  series  and  if 
Mj  +  u^  +  M„  +  •••  is  a  series  such  that  ^  approaches  a  limit  when  n 

«n 

increases,  show  that  the  w-series  converges  absolutely. 


222  CALCULUS 

12.  State  and  prove  a  theorem  for  divergent  series  analogous  to 
that  of  Ex.  IL 

153.  Maclaurin's  series.  It  is  shown  in  algebra  that 
the  quantity  (1  +  x')^,  where  m  is  not  a  positive  integer, 
may  be  developed  into  an  infinite  series  in  powers  of  x  by 
the  binomial  theorem  : 

(1  -\-x)"^=l-\-mx-\ ^-— — ^x^-\ ^^ ^ -or  +  •••, 

the  expansion  being  valid  for  all  values  of  x  numerically 
less  than  1. 

Consider  now  the  problem  of  developing  any  given 
function  f(^x)  in  powers  of  x.  We  will  assume  for  the 
present  that  such  a  development  is  possible,  and  write 

(1)  fXx)  =Cq  +  c^x  +  G^x^  4-  •••  +  c,,x''  +  ..., 
where  the  coefficients  Cq,  c-^,  Cg,  •••  are  constants  to  be  de^ 
termined.      Letting  2j=0,  we  get/(0)  =  c?Q;   i.e.  c^  is  the 
value  of  the  given  function  at  a:  =  0.     Differentiating  (1), 

f\x^  =  Cj  H-  2  ^2^  -h  3  ^32:2  ^  ...^ 
and  setting  2:  =  0,  we  find 

/'(0)=.i. 

Proceeding  in  this  way,  we  find 
/"(0)  =  2.1.„ 
/"_(0)=8_.2.1.3, 

/<»>(0)  =  m!e,„ 

•  •  • 

Hence  (1)  takes  the  following  form,  called  Maclaurins 
series  : 

(2)  /(^)  =  /(0)  +  /'(0):r+-M^+CT):.3+.... 

It  must  be  remembered  that  as  yet  we  have  not  proved 
the  validity  of  this  result ;   we  have  merely  shown  that, . 
if  a  development  in  powers  of  x  is  possible,  it  must  have 
the  form  (2).     Evidently  a  necessary  condition    for  the 


INFINITE  SERIES.     TAYLOR'S  THEOREM        223 

existence  of  Maclaurin's  series  is  that  the  function  and  its 
successive  derivatives  be  defined  at  x  =  0. 

Example  :  Expand  e^  in  Maclaurin's  series. 

Here 

/(^)=6^  hence  /(0)  =  1, 

f"ix)=e^  /"(0)=1, 

•  •  •  • 

•  •  •  •  • 

Therefore  the  development  is 

154.  Taylor's  series.  More  generally,  let  it  be  required 
to  develop  a  function  f(^x')  in  powers  of  x  —  a^  where  a  is 
a  given  number.     Assuming 

/(rr)  =  (?()+  c^{x  —  a)  -\-  c^(x  —  ay  -\-  c^(x  —  ay  +  •••, 
and  setting  x  =  a^  we  find 

/(^)=^o- 
Proceeding  as  in  §  158,  we  obtain  finally  Taylor  s  series: 

(1)  /W=/(a)+/'(<i)(^-<')+-^(Jf-a)' 

Thus  Maclaurin's  series  is  a  special  case  of  Taylor's  series, 
viz.:  the  case  a  =  0. 

When  a  function  is  represented  by  the  series  (1),  a  -e 
say  that  it  has  been  developed  or  expanded  in  Taylor's 
series  about  the  point  x  =  a. 

If,  in  (1),  we  replace  xhj  a  +  h^  we  obtain  another  im- 
portant form  of  Taylor's  series: 

/(a  + A)  =  /(a)  +fXa)h+f^h?+f^^h?+  .... 

It  is  clear  that  the  Taylor  series  for  a  function  f(x^  can 
always  be  formally  written  down  if  the  function  and  its 


224 


CALCULUS 


derivatives  of  all  orders  are  defined  at  rr  =  a.  But  it  by 
no  means  follows  from  this  that  the  series  represents  the 
function  for  any  particular  value  of  x.  The  series  may 
diverge,  or,  if  convergent,  its  sum  may  not  be /(a;).  The 
development  in  Taylor's  series  is  valid  only  for  those  values 
of  X  for  which  the  series  converges  to  the  value  fQc). 

In  the  next  article  we  show  under  precisely  what  cir- 
cumstances a  function  may  be  developed  in  Taylor's  series. 
However,  for  all  functions  that  we  shall  consider,  the 
series,  if  it  converges  at  all,  converges  to  the  value  f(x) ; 
hence  for  those  functions  the  interval  within  which  Taylor's 
series  is  valid  coincides  with  the  interval  of  convergence  of  the 
series. 

Example  :  Expand  the  function  log  x  in  Taylor's  series 
about  the  point  a;  =  1,  and  find  the  interval  of  convergence 
of  the  series. 

In  this  case  a  =  1  : 

fix)  =  log  X,  fay  =  0, 


X 


X' 


/'"W  = 


2_ 


/'(I)  =  1, 

/'"(I)  =  2, 


Hence,  by  (1), 

logx=(x-l}-l(x-  1)2  4-  l(x  -  1)3 

The  general  term  is 

n 

Applying  the  ratio  test,  we  have 

(x  ^  1)"+^ 


lim 

n— >oo 


U, 


=  lim 

n->-oc 


7^4-l 


(x—  ly 
n 


=  la;-l|. 


INFINITE  SERIES.     TAYLOR'S  THEOREM        225 

Thus  the   series    converges  when    |a:  — 1|<1,   i.e.    when 

0  <  2;  <  2.      When  2;  =  0,  the  series  is 

—  1  —  2  —  3  —  •"-, 

which  is  divergent  (§  1-47).      When  x=2,  the  series  is 

1  _  1  4_l 

which  is  convergent  (§  150).     Hence,  finally,  the  interval 
of  convergence  is  0  <  2:  <2. 

EXERCISES 

1.  In  the  Maclaurin  series  for  e*  (§  153),  show  that  the  series 
converges  for  all  values  of  x. 

In  each  of  the  following,  determine  the  interval  of  convergence  of 
the  series. 

2.  Expand  sin  z  in  powers  of  x. 

Ans.   sin  x  =x 1-  —  —  •••.all  values  of  x. 

3 !      5 !  ' 

3.  Expand  cos  x  about  the  origin. 

x^       X* 
Ans.  cos X  =  1 1 •••,  all  values  of  x. 

2!      4! 

4.  Expand  e^  in  Taylor's  series  about  the  point  x  =  2. 

6.   By  replacing  a:  by  1  +  a:  in  the  example  of  §  154,  obtain  the 
development  of  log  (1  +  x)  in  powers  of  x. 

Ans.   log(l  +  a:)=a:-^'+|-^+...,     -l<a:<l. 

6.  Expand  log  (1  —  x)  about  the  origin. 

2  3  4 

Ans.    loga  -x)  =-  X ,  -  1<  a:  <-  1. 

234  '=^ 

7.  Obtain  the  binomial  theorem 

ri    I    ^\m       1    ,           ,  m(m  —  1)    9  ,  m(m  —  1) Cm  —  2)    «  . 
(i  +  a;)"*  =  1  +  mx  -] >^ ^x^  H ^^ ^ ^x^  +  •••. 

2!  3! 

8.  Expand  sin  x  about  the  point  a:  =  -• 

9.  Show  that  log  x  cannot  be  expanded  in  powers  of  x. 

10.  Expand  arctan  x  about  the  origin. 

Ans.    arctan  x  =  x 1 .... 

3       5 

11.  Show  that,  if  P(x)  is  a  polynomial  of  the  n-th  degree  in  x, 

F(x)  =  P(a)  +  P'{a){x  -  a)  +  ^^  {x  -  a)^  +  ...  +  ^^^-li^(x  -  a)", 

2 !  n\ 

whatever  may  be  the  values  of  a  and  x. 

Q 


226  CALCULUS 

155.  Taylor's  theorem.  Let  the  function  f(x)  and  its 
first  n  -{-1  derivatives  be  continuous  in  an  interval  includ- 
ing the  point  a;  =  a,  and  let  a;  =  a  +  A  be  a  second  point  of  the 
interval.  Let  R^  denote  the  difference  between  f(a  +  K) 
and  the  sum  of  the  first  n-{-l  terms  of  the  corresponding 
Taylor's  series  (2),  §  15J: ;  i.e.  set 

(1)  fQa  +  h')=fia)  +  f(a-)h+l^B+- 

n  ! 
For  convenience,  write  R^  in  the  form 

(n  +  1)  I 
so  that 

(2)  /(a  +  A)=/(a)+/'(a)A+  -  +f^-^h'^+ J^P„. 

Consider  now  the  auxiliary  function 

<i>(x)  =fia  +  K)  -/(x-)  -Qa  +  h-  a:)/'(.r) 

{a  +  h-xY  j,,,.^-^       ^^^       (a  +  h-xy  j,^^)^^^ 

2!  n\ 

(a-{-h-xy+'^  p 
(^  +  1)1 

This  function  evidently  vanishes  when  x  =  a  -\-  h,  and,  by 
(2),  it  also  vanishes  when  x  =  a.  Further,  it  results  from 
our  hypotheses  that  (f>C^^  has  a  derivative  <j>'  (x)  in  the  in- 
terval from  x=a  to  x=a-\-h.  Hence  (i>(x)  satisfies  all  the 
conditions  of  Rolle's  theorem  (§  136)  in  that  interval,  and 
its  derivative  must  vanish  at  some  point  x^  of  that  interval. 
Differentiating  <j>(x)^  we  find  after  simplifying  that 

n\  ■ 
By  Rolle's  theorem, 

f(a;i)=0, 
hence 


INFINITE  SERIES.     TAYLOR'S  THEOREM        227 

Substituting  this  value  of  P„  in  (2),  we  get 

(3)  /(a  +  A)=/(a)+/'(a)A  +  -^^^A2+  ...  +/!::^^n 

or,  writing  x—  a  for  A, 

(4)  f{x)=f{a)  +  f{d){x-  a)  +^{x-af-{- ... 

n!     ^         ^         (n+1)!   -^         ^  '^' 
where  a^j  lies  between  a  and  a::. 

P^ormula  (4),  or  its  equivalent  (3),  is  called  Taylor  s 
theorem  tuith  a  remainder.  The  last  term  is  called  the  re- 
mainder after  n  -{-  1  terms : 

(n  H-  1)  1 

For  71=0,  Taylor's  theorem  reduces  to  the  law  of  the 
mean  (§  137): 

f{x-)=fia-)  +  (x-a)f'i^{)- 

If  n  increases  indefinitely,  the  right  member  of  (4)  be- 
comes an  infinite  series,  the  Taylor's  series  iov  f(x).  The 
necessary  and  sufficient  condition  that  the  series  shall  con- 
verge to  the  value /(a:),  and  hence  that  the  function  shall 
be  developable  in  Taylor's  series,  is  that 

^im  R^  =  0. 

Example :  Prove  that  the  function  e*  can  be  developed 
in  powers  of  x  for  all  values  of  x. 

Here  a  =  0  and 

/(a:)  =  e-,  f'Qx-)  =  6^    ••-,  /^"^^>(a:)  =  6% 
so  that  the  remainder  in  Taylor's  theorem  has  the  form 


n+l 


Rn  =  .      .  e-^^ 

(/l+l)I 

where  x^  is  between  0  and  x. 


228  CALCULUS 


^n+l 


By  Ex.  8,  p.  221,  the  quantity  ■- — —  is  the  general 

(w  + 1) ! 

term  of  a  series  which  converges  for  all  values  of  x^  so 

that,  by  §  146,  this  quantity  converges  to  0  when  n  becomes 

infinite.      Hence,  for  all  values  of  x, 

and  the  proof  is  complete. 

EXERCISES 

1.  In  the  Maclaurin  series  for  sin  x,  prove  that  the  remainder  con- 
verges to  0  for  all  values  of  x. 

2.  Prove  that  cos  x  can  be  expanded  in  powers  of  x  for  all  values 
of  X. 

156.  Approximate  computation  by  series.  We  have 
found  in  the  preceding  article  that  any  function  whatever, 
provided  certain  conditions  regarding  continuity  are 
satisfied,  can  be  represented  by  Taylor's  theorem  as  a 
polynomial  of  arbitrary  degree,  with  a  certain  remaiyider 
Rn-  It  is  clear  that  i2„  is  the  error  committed  if  we  re- 
place the  function  by  the  polynomial. 

This  suggests  a  method  for  computing  approximately 
the  numerical  value,  for  any  given  value  of  the  argument, 
of  functions  such  as  the  sine,  cosine,  logarithm,  etc.,  whose 
value  cannot  be  found  directly.  We  have  only  to  build 
up  the  Taylor  polynomial  for  the  function  in  question, 
and  show  that  the  error  R^  is  less  than  the  allowable  limit 
of  error  for  the  problem  in  hand. 

If  now  our  function  can  be  developed  in  Taylor's  series, 
we  know  at  once  that  its  value  to  any  desired  degree  of  ac- 
curacy can  be  found  by  merely  adding  up  a  sufficient  num- 
ber of  terms  at  the  beginning  of  the  series.  For,  by  §  155, 
the  remainder,  or  error,  R^  converges  to  0  as  ti  increases 
indefinitely,  and  hence,  by  the  definition  of  §  14,  can  be 
made  as  small  as  we  please  by  taking  n  sufficiently  large. 

An  upper  limit  for  the  error  committed  by  stopping  at 


INFINITE  SERIES.     TAYLOR'S  THEOREM       229 

any  point  may  frequently  be  found  from  the  general  prop- 
erties of  series.  Thus  in  the  case  of  an  alternating  aeries 
the  error  is  less  than  the  first  term  neglected,  by  theorem 
II  of  §  150. 

Example  :    Compute  sin  3°  correct  to  five  decimal  places. 

Since  ^       ^ 

sin:r=2:-— +--••., 

it  follows  that 


sin3°  =  sin^  =  — -i  —     +         ,       , 
60      601     6V60y       120V60y 

=  0.052365-0.000024+  •••. 

Since  this  is  an  alternating  series,  the  error  committed  by 
stopping  with  any  term  is  less  than  the  next  term.  With- 
out computing  the  third  term,  we  see  that  it  is  much  too 
small  to  affect  the  fifth  decimal  place,  hence  we  need  keep 

only  two  terms : 

sin  3°  =0.05234. 

To  be  of  practical  use  in  computation,  a  series  should 
converge  rapidly,  as  in  the  above  example,  so  that  a  few 
terms  are  enough  to  give  the  desired  degree  of  accuracy. 
In  this  connection  the  following  point  should  be  noted. 
Our  choice  of  a  in  Taylor's  theorem  is  governed  only  by 
the  necessity  of  knowing  at  that  point  the  value  of  /C^:) 
and  its  derivatives.  Since  the  remainder  Mn  contains  the 
factor  (x— a)"+\  it  is  clear  that,  in  general,  the  smaller  the 
difference  x—  a^  the  faster  the  remainder  will  approach  0. 
Hence  in  general,  of  all  possible  values  for  a,  we  should 
choose  that  one  lying  nearest  to  the  value  of  x  in  question. 
Thus  to  compute  sin  3°,  we  took  a  =  0 ;   if  we  had  to  com- 

pute  sin  47°,  we  would  take  *  a  =  — ,  i.e.  we  would  expand 

•i 

sin  x  in  powers  of  2; ;   etc. 

4 

*  Assuming  of  course  that  we  know  the  value  of  the  sine  and  cosine 

only  for  the  "principal  angles  '0,  — ,  -,  etc. 

6     4 


230  CALCULUS 

EXERCISES 

1.  Draw  on  the  same  axes,  on  a  large  scale,  the  curve  y  =  sinx 

and  the  first  and  second  "  approximation  curves  "  y  =  x,  y  =  x  —  —  ^  in 

6 

the  interval  0  <  a;  <  tt.     Estimate  the  interval  within  which  each  of 
the  approximating  polynomials  is  correct  to  one  decimal  place. 

2.  Proceed  as  in  Ex.  1  with  the  curve  ?/  =  e^  and  the  successive 

approximation  curves  y  =  1  -hx,  y  =  l  +  x  +  —  ,y  =  l  +  x-\ 1 —  ,in 

the  interval  -  2  <  a:  <  2. 

3.  Compute  (a)  sin  1°  to  five  places;  (h)  sin  9°  to  three  places; 
(c)  cos  3°  to  four  places. 

4.  Find  the  value  of  e  to  five  decimal  places.  Ans.  e  =  2.71828  •••. 

5.  Compute  (a)  sin  47°,  (b)  cos  31°,  each  to  four  places. 

6.  Find  the  tenth  root  of  e  to  five  decimal  places. 

7.  Find  the  value  of  e^-^^  to  five  decimal  places. 

8.  Show  that  an  arc  of  a  great  circle  of  the  earth  2|  miles  long 
recedes  1  ft.  from  its  chord. 

9.  Taking  the  circumference  of  the  earth  as  40,000,000  meters, 
show  that  the  difference  between  the  circumference  and  the  perimeter 
of  a  regular  inscribed  polygon  of  1,000,000  sides  is  less  than  one  fif- 
teenth of  a  millimeter. 

10.  Within  what  interval  can  sin  6  be  replaced  by  0,  if  accuracy 
to  three  decimal  places  is  required  ? 

157.  Operations  with  power  series.  Operations  that  can 
always  be  performed  upon  series  of  a  finite  number  of 
terms,  such  as  rearrangement  of  terms,  multiplication  of 
one  series  by  another,  term-by-term  differentiation  or 
integration,  etc.,  cannot  be  assumed  offhand  to  be  allow- 
able with  infinite  series,  and  in  fact  it  is  easily  shown  that 
they  are  not  allowable  in  all  cases. 

In  dealing  with  developments  in  Taylor's  series,  it  is 
frequently  desirable  to  know  just  when  such  elementary 
operations  are  permissible.  We  therefore  state,  without 
proof,  the  following  theorems  regarding  power  series. 

Theorem  I :  Addition.  Two  poiver  series  may  he  added 
together  for  all  values  of  x  for  which  both  series  are  convergent. 


INFINITE  SERIES.     TAYLOR'S  THEOREM       231 

That  is,  if  the  series 

</>(a;)  =  (Iq  +  a^x  +  a^x^  +•••■, 

are  both  convergent,  the  series  obtained  by  adding  these 

together  will  converge  to  the  value  <^(a:)  +  '^(x)  : 
(j)(x)  H-  yjr^x)  =  a^  -f  6q  +  (aj  +  ^i):^^  +  (^2  +  ^2)^  +  •*•• 
Theorem    II  :    Multiplication.       Two  power   series 

may  he  multiplied  together  for  all  values  of  x  for  which  both 

series  are  absolutely  convergent. 
That  is,  if  the  series 

(f>(x)  =  ^0  +  a^x  H-  a<^x^  +  •••, 
-^(x)  =  ^0  +  ^1^  +  M^  +  ••• 

are  both  absolutely  convergent,  then 

(f>(x)  -^(x)  =  a^b^  H-  (^i^o  +  «o^i)a: 

+  («2^o  +  ^1^1  +  <^oh)^^  +  •  •  •  • 

Theorem  III:  Division.  One  convergent  power  series 
may  be  divided  by  another.,  provided  the  constant  term  in  the 
denominator  is  not  0.  The  result  holds  within  a  certain 
interval,  the  determination  of  which  is  beyond  the  scope 
of  this  discussion. 

Theorem  IV  :  Substitution.     If  the  series 
z  =  a^^-\-  a^y  -{■  a^y'^  +  ••• 
converges  for  all  values  of  y,  aiid  the  series 
y  =  bQ-}-b-^x  +  b^x^  +  ••• 
converges  for  all  values  of  x,  the  seizes  for  y  may  be  sub- 
stituted in  the  series  for  z  and  the  result  arranged  in  poivers 
of  X.     This  result  holds  for  all  values  of  x. 

Theorem  V :  Differentiation.  A  poiver  series  may 
he  differentiated  term  hy  term  for  all  values  of  x  within  its 
interval  of  convergence.* 

Theorem  VI  :  Integration.  A  poiver  series  may  he 
integrated  term  by  term  between  any  limits  lying  within  the 
interval  of  convergence.* 

*  The  endpoints  of  the  interval  are  excluded. 


232  CALCULUS 

These  theorems  enable  us  to  obtain  many  Taylor  ex- 
pansions in  which  the  evaluation  of  the  successive  deriva- 
tives would  be  very  tedious,  and  in  which  the  law  of 
formation  of  the  coefficients  is  so  complicated  that  the 
interval  of  convergence  could  not  be  determined  directly. 

Example :  Expand  e^^^^  in  powers  of  a;,  to  x^  inclusive. 

By  Exs.  1,  2,  p.  225,  we  have 

sin^      -,    ,     .         .   sin^a^  ,  sin^a:  ,    sin*a;  , 


21  3!  4! 


sin  ic  =  2^  —  —  + 


Since  both  these  series  converge  for  all  values  of  a:,  the 
series  for  sin  x  may,  by  theorem  IV,  be  substituted  in  the 
series  for  e^^^^: 

.-=i  +  ^,_|i  +  ...)+(|_,_|!  +  ...J 

^1/  3;8  N3        1   /         a?   ,       Y_^ 

+  3l("-3T+-)   +r!  I" -3! +  •••;■*"•••• 

Expanding  the  parentheses  by  theorem  II  and  collecting 

terms,  we  find 

By  theorem  IV,  this  series  converges  for  all  values  of  x. 

EXERCISES 

Expand  the  following  functions  in  powers  of  x,  and  determine  the 
interval  of  convergence  in  each  case. 

1.  sin^a:.  Ans.   x^  -  |  a:*  +  53  x^  +  •••,  all  values. 

2.  cos^o:.     Compare  this  result  with  that  of  Ex.  1. 

•3  5 

3.  e'^sin  x.  Ans.  x +x^ -{-—-  — +  ••• ,  all  values. 

3      30 

4.  log  i^il£.  6.   e=«<=08*. 

1  —  X 

6.    By  integrating  the  series 

=1— X+X^  —  X^  +  --' 

1  +  X 

between  the  limits  Oandx,  obtain  the  Maclaurin  series  for  log  (1  +  x). 


INFINITE  SERIES.     TAYLOR'S  THEORExM       233 

7.    Expand  arcsin  x  in  powers  of  x  by  integrating  the  binomial 

f         1 
expansion  of  . 

/8.    Expand by  division,  and  integrate  the  resulting  series 

^  1  +  x"-^ 

term  by  term.     Cf.  Ex.  10,  p.  225. 

9.    By  differentiating  the  Maclaurin  series  for  log  (1  —  x)  (Ex.  6, 

p.  225),  prove  the  formula  of  elementary  algebra  for  the  sum  of  an 

infinite  geometric  series. 

10.  By  means  of  series,  prove  the  formula  sin  2  x  =  2  sin  x  cos  x. 
Expand  each  of  the  following  in  powers  of  a:;  the  interval  of  con- 
vergence is  as  indicated  in  each  case. 

11.  -^.  -l<x<l. 
1  —  X 

12.  sec  X.  Jns.  1 +-^2  + ^a;4+ ...,    _?<x<- 

2  24  2  2" 
1             o                           _  _ 

13.  tana;.  Ans.  x -]- -x^ -^ —x^  ^  •■■,     -^^x<'- 

3  15  2  2 

14.  Show  that,  for  values  of  x  so  small  that  the  fourth  and  higher 

wers  of  —  may  be  ne^ 
a 

replaced  by  a  parabola. 

15.  Find  the  area  under  the  curve  y  —  from  x  =  0  to  x  =  \. 

Draw  the  figure.  Ans.  0.9461. 

16.  Find  the  centroid  of  the  area  in  Ex.  15. 

1     9 

17.  Find  the  area  under  the  curve  y  =  e    -  ,  from  x  =  0  to  z  =  1. 

Ans.  0.86. 

18.  The  area  in  Ex.  17  revolves  about  the  a:-axis.  Find  the 
centroid  of  the  volume  generated. 

19.  Raise  1.03  to  the  fifth  power.  Ans.  1.16. 

20.  Show  that,  in  leveling,  the  correction  for  the  curvature  of  the 
earth  is  8  in.  for  one  mile. 

21.  A  mountain  peak  1  mile  high,  situated  on  an  island,  is  just 
visible  from  the  mainland.  If  there  is  no  refraction,  how  far  out  at 
sea  is  the  island?     Solve  in  two  ways.  Ans.  90  miles. 

22.  Show  that  the  length  of  the  arc  of  the  hyperbola  xj/  =  1  from 
x  =  100  to  a:  =  200  differs  from  the  length  of  the  chord  by  one  part  in 
5,000,000,000. 


powers  of  -  may  be  neglected,  the  catenary  y  =  -  (e"  +  e   ")  may  be 


234  CALCULUS 

23.  Find  the  difference  between  the  circumference  of  the  earth  and 
the  perimeter  of  a  regular  circumscribed  polygon  of  1,000,000  sides. 
Cf.  Ex.  9,  p.  230. 

24.  Find  the  surface  generated  by  revolving  the  curve  y  =—  about 
the  X-axis,  from  a;  =  0  to  a:  =  ^. 

158.    Computation  of  logarithms.     We  have  found 

/y^  /yiO  /y**x 

log(l+a;)=a;--+-- j+  •••, 


/vi**  /v»t>  /y»*t 

log  (\  —  X)  =  -~  X 


each  series  holding  for  values  of  x  numerically  less  than  1. 
From  these  vre  may  deduce  a  series  that  is  better  adapted 
to  numerical  computation  than  either  of  the  above  series. 
Subtracting  the  second  equation  from  the  first  (by 
theorem  I,  §  157),  we  find 

1       l-\-x      ^(     ,  x^  .  x^  , 

for  values  of  x  numerically  less  than  unity.     Let  us  put 

1  +  ^_  "^  +  1 
1  —  X         m 

or 

1 

x  = 


2m-f  l' 

where  m  may  have  any  positive  value.     Then 
(1)   log  (1  +  m)  =  log  m-\-2 


1    +      1 


2  m -hi      3(2  m  + 1)3 

+ L_.  .. 

5(2^  +  1)5 

This  series  converges  rapidly,  and  is  therefore  well 
adapted  to  computation.  It  is  easily  shown  that  for 
values  of  m  >  1  the  error  committed  by  stopping  at  any 
point  is  only  slightly  greater  than  the  first  term  neglected. 


INFINITE  SERIES.     TAYLOR'S  THEOREM       235 
Example  :  Taking  m  =  1,  we  have 


log2  =  2g 


3.33^5.35^7.3' 


=  2[0.3333  +  0.0123  +  0.0008  +  .••] 
=  0.693. 

From  this  the  logarithms  of  4,  8, ...  may  be  found  directly. 
With  m  =  2,  we  find  log  3  ;  from  this  and  the  previous 
result  we  may  obtain  the  logarithms  of  all  numbers  whose 
only  prime  factors  are  2  and  3.  In  fact,  it  is  clear  that 
only  the  logarithms  of  prime  numbers  need  be  computed 
by  the  series. 

EXERCISES 

1.  Compute  to  three  decimal  places  the  natural  logarithms  of  all 
integers  from  3  to  10  inclusive. 

2.  After  finding  log  10,  obtam  logj^  2,  logjQ  3. 

3.  By  comparison  with  the  geometric  series 

1  1  Ti    ,  1  ,  1 


2n  +  l     (2  7n  +  l) 


fl   I  1  1  1  ,    ...1 

^«+iL        (2  m  +  1)2      (2  7n  +  1)4  J' 


prove  that  the  error  committed  by  stopping  with  the  n-th  term  of  the 

1 
1 
(2  w  +  1)2 


series  (1)  is  less  than  times  the  first  term  neglected 


CHAPTER    XXI 

FUNCTIONS  OF   SEVERAL  VARIABLES 
I.    Partial  Differentiation 

159.  Functions  of  several  variables.  Up  to  this  point 
we  have  been  concerned  with  functions  of  a  single  argu- 
ment. A  function  may,  however,  depend  upon  several 
independent  variables.  For  example,  the  volume  of  a  cir- 
cular cylinder  is  a  function  of  its  radius  and  altitude  ;  the 
acceleration  of  a  moving  particle  is  a  function  of  all  the 
forces  acting  on  it ;  the  strength  of  a  rectangular  beam 
is  a  function  of  its  breadth  and  depth. 

If  2  is  a  function  of  two  variables  x  and  y,  we  write 

with  a  similar  notation  for  functions  of   more  than  two 
variables. 

Geometrically  a  function  of  two  variables  may  be  rep- 
resented as   the   ordinate   of  a   surface    in   space.     Thus 

the  equation 

z  =  ax  -\-  hy  -\-  c 

represents  a  plane  ;  the  equation 

z^Q?  —  if' 
represents  a  hyperbolic  paraboloid,  etc. 

A  thorough  study  of  functions  of  several  variables  is 
beyond  the  scope  of  a  first  course  in  the  calculus.  In  the 
present  chapter  we  set  forth  a  few  of  the  most  important 
definitions  and  theorems,  confining  our  attention  chiefly 
to  functions  of  two  arguments. 

160.  Limits ;  continuity.  Suppose  we  have  given  a 
function  of  two  variables 

(1)  2 = /(^,  y) 

236 


FUNCTIONS  OF  SEVERAL  VARIABLES  237 

representing  a  surface  in  space.  When  x  and  y  approach 
the  respective  values  x^^  ?/q,  the  function  z  is  said  to 
approach  a  limit  Zq  if  the  point  (2;,  y,  2)  of  the  surface  (1) 
approaches  a  definite  limiting  point  (a^^,  y^,  Zq).  In  other 
words,  if  when  x  is  sufficiently  near  Xq  and  y  is  sufficiently 
near  y^  the  difference  between  z  and  Zq  becomes  and  re- 
mains numerically  less  than  any  preassigned  quantity 
however  small,  then  z  is  said  to  approach  the  limit  Zq  :  in 

symbols, 

lim  f(^x,  y)  =  z^. 

A  function /(a;,  y)  is  said  to  be  continuous  at  the  point 
\\mf(x,  y)  =  f(xQ,  y^). 

X—^Xq 

Similar  definitions  are  laid  down  for  functions  of  more 
than  two  variables. 

In  what  follows,  it  is  supposed  that  all  functions  occur- 
ring are  continuous  at  all  points  under  consideration. 

161.  Partial  derivatives.  If  y  be  kept  jixed^  the  func- 
tion ^,       , 

becomes  a  function  of  x  alone,  and  its  derivative  may  be 
found  by  the  ordinary  rules.  This  derivative  is  called 
the  partial  derivative  of  z  with  respect  to  a;,  and  is  denoted 
by  any  one  of  the  symbols 

ax      dx 

The  partial  derivative  with  respect  to  y  has  a  similar 
meaning. 

The  idea  of  partial  differentiation  may  be  extended  at 
once  to  functions  of  any  number  of  variables.  We  have 
only  to  remember  that  in  differentiating  with  respect  to 
any  one  variable,  all  the  other  variables  are  treated  as 
constants. 


238 


CALCULUS 


Fig,  84 


162.  Geometric  interpretation  of  partial  derivatives.     To 

keep  y  constant,  say  y  =  y^^  in  the  equation 

means  geometrically  that  we  cut 

the  surface    by    the  plane   y  =  ^q. 

.    ■    dz  . 
The  partial  derivative  —  is  tliere- 

fore  the  slope  of  the  curve  of  in- 
tersection of  the  surface  and  the 
plane,  i.e.  of  the  curve  whose  equa- 
tions are 

The  partial  derivative  —  may  be 

interpreted  similarly.        ^ 

3z      dz 

163.  Higher  derivatives.     The  derivatives  —-,  — -   are 

*  ydxdy 

themselves  functions  of  x  and  y,  and  their  partial  deriva- 
tives can  in  turn  be  found.  They  are  denoted  by  the 
following  symbols : 

by  \dxj      by  ox 

dx\dyj      dxdy       ""^  ,  . 

dy\dyj      dy^ 
The   process  can  of   course   be    repeated   to   find  still 
higher  derivatives. 

It   can   be   shown    that    the   two    "cross-derivatives" 

5%    dh 


dydx     dxdy 


are  identical : 


d^z 


dydx      dxdy 


FUNCTIONS  OF  SEVERAL  VARIABLES  239 

That  is,  the  order  of  differentiation  is  immaterial.  This  is 
true  for  derivatives  of  all  orders,  and  for  functions  of  any 
number  of  variables. 

EXERCISES 

•J       <% 

Find  -T- ,  ^—  for  the  following  functions. 
ox    dy 

1.  z  =  x^  -\-  xy  —  S  X  -{-  .5.  Ans.    —  =  2  x  +  y  —  S ',  ■—  =  x. 

dx  oy 

2.  2  =  a:^  +  3  x'^y  —  xy  -\-  2  y  —  3. 

3.  2  =(x2  -  2  xyy.  ^.    z=  ^^^~^y^ . 


5.   2  =  ecos^cx-y)^  6.   2  =  logVa:2  +  3^2. 

«  .        y  A  dz         —  y 

7.  z  =  arctan  --^  •  ^  "^s.    —  =  — — ^ . 

a;  5a;      a;^  +  ?/^ 

8.  Given /(a:,  y,  2)  =  xyz  +  3  .r^?/  +  z^,  find/,,  /j^,/,. 

-4  ns.  /,  =  y(z  +  6  a:). 

9.  It  u  =  x^  +  y^  —  z^  +  X  +  2  y,  find  — ■ ,    — ,  —  • 

dx     oy     oz 

10.  Find  the  slope  of  the  curve  cut  from  the  hyperbolic  paraboloid 
2  =  a:2  —  2^2  by  the  plane  ?/  =  3,  at  the  point  (4,  3,  —  2).  Ans.   8. 

11.  Find  tlie  equations  of  the  tangent  to  the  parabola 

2  =  3  x2  +  4  ?/2,  a;  =  2 
at  the  point  (2,  1,  16).  Ans.   x  =  2,z=Sy^^. 

12.  If  M  =  x^y-  -  2  xy^  +  3  x^y^,  show  that  x-:^  +  y~-bu. 

13.  If   w  =  (?/  -  z){z  -  x)(^x  -  y),  show  that    ^  +  -^  +  -^  =  0. 

ox      oy      oz 

(92^  fi-2z  S-"  52- 

14.  Given  2  =  a:  +  a;S^2  +  2  xY,  find  ^,    -^,    -^,    ^  • 

oa;2     ayaa;     aa;a_y     oy^ 

^•2-y  Q'2-, 

A  ns.   — -  =  6  a;?/2  +  4:y*;        '^    =  6  a-2?/  +  16  xy^. 
dx^  oyox 

d'-~         52- 

15.  Given  2  =  a:2^2  _j_  3  ^^a^^s  _  ^2^^  verify  that 


dydx      dxdy 

d^z  5^2     d^z 

16.  If  2  =  cos  (a:  —  y),  show^  that      =       =  - — —  • 

ayox^      oxoyox      ox'-oy 

17.  If  M  =  ei-2/-2^,  verify  that 

52^   _   52^  .     d'^u  _  d'^u 
dxdy      dydx     dydz      dzdy 


240  CALCULUS 

18.  If  ^  =  i  log  {x^^-y^),  show  that  |^  +  ^  =  0. 

19.  li  z  =  x'^y,  show  that  x—-  +  ^/-^  =  02. 

^  dx        dy 

20.  If  M  =  x^+  y^+  y^,  show  that  x^  +  y^  +  zp:  =  2u. 

dx        dy        dz 

21.  lif(x,  y,  z)  =  ^  show  that/^2  +/^2  +/,,  =  0. 

Vx^  +  ^/'-^  +  z'^ 

22.  Prove  that  if  two  functions  u  and  y  are  so  related  that 

du  _  dv       du  _      dv 
dx     dy      dy         dx 

then  '  d^^d^  =  o. 

dx^      dy'^ 

164.    Total   differentials.     When   x  and    ?/  change   by 
amounts  Ax  and  A^,  the  function 

z=f(x,^} 

changes  by  an  amount  Az.     It  can  be  shown  that  Az  may 
be  expressed  in  the  form 

dz  dz 

Az  =  —Ax-\ Ay  +  eAx  +  riAy, 

dx  dy 

where  €  and  7]  are  infinitesimals. 

The  quantity  — Ax-\ Ay  is  called  the  principal  part 

dx  dy 

(cf .  §  49)  of  the  infinitesimal  Az.     The  total  differential  of 

z  is  defined  as  the  principal  part  of  Az : 

J        dz  .       .   dz  . 
dz  =  — Ax-\ Ay. 

dx  dy 

In  particular,  ii  z  =  x,  —  =1  and  —  =  0,  so  that 

dx  dy 

dx  =  Ax. 
Similarly 

dy  =  Ay. 
Hence  we  may  write 

(1)  dz  =  ^dx  +  ^^dy. 

dx  dy 


FUNCTIONS  OF  SEVERAL  VARIABLES  241 

For  functions  of  more  than  two  arguments  a  similar 
formula  holds.     Thus,  if 

(2)  du  =  ^—dx  -\ dy  -\ dz. 

ox  dy  dz 

If  x^  y,  z  are  functions  of  a  fourth  variable  f,  then  u  be- 
comes a  function  of  t  alone,  and  its  differential  has  been 
defined  in  §  50.  It  can  be  shown  that  the  value  of  du 
as  given  by  (2)  agrees  with  the  earlier  definition,  so  that 
(2)  still  holds  even  when  x,  y,  z  are  functions  of  a  single 
variable. 

Example:  Find  approximately  the  increase  in  the  area 
of  a  rectangle  if  each  of  its  dimensions  increases  by  a 
small  amount. 

We  have 

A  =  ah^ 
hence 

dA  =  hda  -|-  adh. 

The  actual  increase  in  the  area  is 

^A=(a-\- da)(h-\-dh)—ah 
=  hda  -f  adh  +  dadh. 

If  da  and  dh  are  so  small  that  their  product  can  be 
neglected  in  comparison  with  the  other  terms  occurring, 
the  total  differential  dA  represents  the  actual  change  AA 
with  sufficient  accuracy. 

165.  Differentiation  of  implicit  functions.  Let  y  be  de- 
fined as  a  function  of  x  by  the  equation 

K^,  2/)  =  0. 
Let  us  for  an  instant  put 

then  by  (1),  §  164, 

dz  =  ^-dx  -[-^-  dy, 
dx  dy 


da 

bda 

(5 

a 

Si- 

I 

db 

df 

dy 

dx 

dx 

df 

9y 

242  CALCULUS 

But  in  the  present  instance 

2  =  0, 

hence 

df  f 

dz  =  -^  dx  -\-  ^^  d^  =  0, 
dx  dy 

or 

(1)  -£=-"^  (1^0). 


The  value    of  — ^   as  g^iven  by  this  formula  is  of  course 
dx        ^  -^ 

identical  with  that  given  by  the  method  of  §  25. 

Again,  let  z  be  defined  implicitly  as  a  function  of  the 

two  independent  variables  x  and  y  by  the  equation 

F(:x,y,z)=0. 

Put 

u  =  F(x,  y,  z)  ; 
then 

,        dF.     ,  dF.     ,  dF. 

du  =  — -dx+  -—-ay  -\ dz, 

dx  dy  dz 

But  since  t^  =  0,  du=  0  likewise,  and 

—  dx-\ dy  -\ dz  —  0. 

dx  dy  dz 

Further,  since  s  is  a  function  of  x  and  y,  we  may  write 

dz  dz 

dz  =  —  dx  -\ dy. 

dx  dy 

Eliminating  dz  between  these  two  equations,  we  find 

dx        dz  dx]  \dy        dz  dy) 

To  find  — ,  keep  y  fixed,  so  that  dy  =  0.     Then 
dx 

dF     dFdz^^Q 

dx       dz  dx 


FUNCTIONS  OF  SEVERAL  VARIABLES  243 


or 


dF 

dz 

dx 

dx 

dF 

dz 

dF 

dz 
dy 

dy 
dF 

dz 

Similarly 


(3)  £=-.^  {~*0 


EXERCISES 

Find  the  total  differential  of  each  of  the  following  functions. 
1.  z  =  x^  -  'S  xy  +  y-  +  2  y.  2.   z  =  cos^  (x  -  y). 

3.   u  =  X  -\-  y  +  z.  ^.   u  =  log  tan  ^' 

X 

5.  Let  V  be  the  volume,  S  the  total  surface,  of  a  right  circular 
cylinder.  If  r  and  h  change  by  an  amount  Ar  and  A^  respectively, 
find  d  V,  A  V,  dS,  A5.     Draw  a  figure. 

6.  In  Ex.  5,  if  r  =  5  ft.,  A  =  10  ft.,  Ar  =  AA  =  2  in.,  compute  the 
percentage  of  error  made  by  using  dV  in  place  of  AF  and  dS  in  place 
of  A5. 

7.  The  dimensions  x,  y,  z  oi  a  rectangular  parallelepiped  change 
by  amounts  Aa:,  Ay,  Az.  Find  dV,  AV.  Also  obtain  dV  and  AF 
directly  by  inspection  of  a  figure. 

Find  -^  in  the  following  cases,  using  formula  (1)  of  §  165. 

8.  3x-4ty  +  2xy  =  \. 

9.  (2x2-  3?/2)2+  1  _  a;^^o. 

10.  Arctan  ^  =  x. 

X 

11.  Given  x^  +  y^ -\- z^  =  1,  find  ^,  ^  by  formulas  (2)  and  (3) 

dx    dy  ■ 
of  §  165. 

12.  Find  the  equations  of  the  tangent  to  the  circle  x'2-i-y^  +  z^=d6, 
y  =  4'at  the  point  (2,  4,  4). 

13.  Find  the  equations  of  the  tangent  to  the  ellipse  ar^  +  3  3/2  =  z^, 
s  =  4  at  the  point  (2,  2,  4). 


244  CALCULUS 

II.    Applications  to  Solid  Analytic  Geometry 

166.   Tangent  plane  to  a  surface.     It  can  be  shown  that 
all  the  lines  tangent  to  a  surface 

z  =  f{x,  y) 

at  a  point  P  :  (a^Q,  ^q,  Zq)  lie  in  a  plane,*  the  tangent  plane 
to  the  surface  at  that  point.  This  plane  is  of  course 
determined  by  any  two  of  the  tangent  lines.  We  have 
already  learned  (§  162)  how  to  find  the  equations  of  the 
tangent  lines  lying  in  the  planes  x  =  Xq^  y  =  y^.  Let  us 
assume  the  equation  of  the  tangent  plane  in  the  form 

2  -  2^0  =  %(^'  -  ^o)  +  ^2(^  -  ^o)' 

where  m^  and  m^  are  to  be  determined.     Now  the  line  of 

intersection  of  this  plane  with  the  plane  y  —  y^  has  the 

slope  my     But  this  line  is  the  tangent  lying  in  the  plane 

bz 
y  =  ^Q,  and,  by  §  162,  its  slope  is  the  value  of  ^^  at  P, 


Bz 
which  value  we  shall  denote  by  the  symbol 


^      dx 


dxjp 

dz 


Hence 


Similarly  we  find 

^z-^ 
^y\p 

Thus  the  equation  of  the  plane  tangent  to  the  surface 

dz 


(1)  ^~^o  =  ^     (^-^o)  +  i5 


p 


iy-yo)' 


dyl 

More  generally,  let  the  equation  of  the  surface  be  given 
in  the  implicit  form 
(2)  FCx,y,z)  =  0, 

*  Provided  s,  —  ,  and  ^  are  continuous  at  P. 
dx  dy 


FUNCTIONS  OF  SEVERAL  VARIABLES 


245 


where  the  partial  derivatives^ — ,  — ,  ^— do  not  all  vanish 

dx     dy     dz 

at  P  :  (xq,  t/q,  Zq).     Suppose  for  definiteness  that  -—    ^0. 

dz  _Jp 

We  may  imagine  equation  (2)  solved  for  z,  and  may  then 

write  the  equation  of  the  tangent  plane  by  formula  (1). 

But,  by  (2)  and  (3)  of  §  165, 

dF  dF 

dz___^     i£^_^. 
ex~      dF'    By  dF 

dz  dz 

Substituting  these  values  in  (1),  we  find 


BF; 

dx 


'dz 


dF 

dy  _ 


jp 


(x  —  Xq)        Q-p 

~dz 


iy  -  Vo)^ 


or 


^  ^    fix 


p  dy_ 


(j^-j/„)+g"' 

p  dz 


(z  -  z.)  =  0. 


167.  Normal  line  to  a  surface.  The  normal  to  a  surface 
at  a  point  P  is  the  line  through  P  perpendicular  to  the 
tangent  plane. 

It  will  be  recalled  from  solid  analytic  geometry  that  the 
direction  cosines  of  any  line  perpendicular  to  the  plane 

Ax  +  By  +  Cz  +  D=0 

are  proportional  to  the  coefficients  A^  B,  C.  Hence,  since 
the  normal  is  perpendicular  to  the  tangent  plane  (3)  of 
§  166,  we  have  at  once  the  following 

Theorem:  The  direction  cosines  of  the  normal  to  the  sur- 
face 

F(x,y.z)=0 

at  any  point  are  proportional  to  the  values  of  — ,  — ,  —  at 
,7    ,       .  ^  dx     dy     dz 

that  point.  ^ 

This  theorem  is  fundamental  in  the  geometry  of  surfaces. 


246  CALCULUS 

By  analytic  geometry,  the  equations  of  a  line  through 
%  ^0'  ^0  with  direction  cosines  proportional  to  a,  5,  c  are 

^  ~  ^0  _  ^  ~  l/o  _  ^  ~  ^0. 

From  this  the  equations  of  the  normal  at  any  point  may  be 
written  down  at  once. 

168.  Angle  between  two  surfaces ;  between  a  line  and  a 
surface.  The  angle  between  two  surfaces  at  a  point  of  in- 
tersection is  defined  as  the  angle  between  the  tangent  planes 
at  that  point,  and  this  in  turn  is  equal  to  the  angle  be- 
tween the  normals.  This  angle  may  be  found  by  the 
theorem  of  analytic  geometry  that,  if  two  lines  have  direc- 
tion cosines  Zp  m^,  n^  and  l^,  m^,  n^  respectively,  the  angle 
between  them  is  given  by  the  formula 
cos  6  =  l^l^  +  m^rric^  +  n^n^- 

The  angle  at  which  a  line  pierces  a  surface  is  defined  as 
the  angle  between  the  line  and  the  tangent  plane  at  the 
piercing-point.  This  is  evidently  the  complement  of  the 
angle  between  the  line  and  the  normal. 

Example  :  Find  the  angle  between  the  cylinder  ?/2  =  4  a; 
and  the  ellipsoid  2x^  -\-  g^  +  z^  =  1  ?Lt  the  point  (1,  2,  1). 

For  the  ellipsoid,  the  partial  derivatives  are 

dF      .       dF     ^       dF     ,-, 

—  =4  a;,   —  =  ^J/^   —  =  2  z, 
dx  By  dz 

hence  the  direction  cosines  of  the  normal  at  (1,  2,  1)  are 
proportional  to  4,  4,  2,  and  their  actual  values  are  J,  |,  J. 
For  the  cylinder, 

dX  dg  dz 

hence  the  direction  cosines  of  the  normal  are n,  — -<>  0. 

Therefore  "^2    V2 

cos(9  =  -  — .?  +  — .?=0: 
V2     3      V2     B 

the  surfaces  intersect  at  right  angles. 


FUNCTIONS  OF  SEVERAL  VARIABLES  247 

EXERCISES 

Find  the  equations  of  the  tangent  plane  and  normal  line  to  each  of 
the  following  surfaces  at  the  point  indicated. 

1.  The  cone  x-  + 'd  if  =  z^  dit  (2,  2,  4) ;  draw  the  figure. 

Ans.    x+^y  -2z  =  0;   -^-^  =  ^Ln^  ^  ^jui . 
^  '13-2 

2.  The  paraboloid  z  —  x^  —  ?/'^  at  (1,  1,  0). 

3.  The  cylinder  y^  =  4:  ax  at  (a,  2  a,  «) ;  draw  the  figure. 

4.  The  paraboloid  x  =  yz  a,t  the  origin. 

5.  The  sphere  x^  -\-  y^  -{-  z^  —  a^  at  {xq,  y^,  z^). 

Ans.    x^x  +  y^y  +  zqz  =  a^. 

6.  The  surface  — ±f-±-=lat  (a^o,  yo,  Zq). 

7.  Find  the  equations  of  the  tangent  to  the  circle  x'^  -\-  y'^  -\-  z^  =  ^, 
x  +  y-\-z=^^t  the  point  (1,  2,  2)  ;  draw  the  figure. 

8.  Find  the  angle  between  the  sphere  x^  +  y^  +  s;^  =  14  and  the 
ellipsoid  3  a;^  +  2  ^2  ^  s^  ^  20  at  the  point  (-1,  -  2,  3).  Ans.  23°  33'. 

9.  Show  that  at  any  point  on  the  2-axis  there  are  two  tangent 
planes  to  the  surface  a^y'^  =  x^(b'^  —  z^) . 

10.  Show  that  the  sum  of  the  squares  of  the  intercepts  on  the  axes 

2  2  2  2 

made  by  a  tangent  plane  to  the  surface  x^  +  y'^  -{■  z'^  =  a^  is  constant. 
Sketch  this  surface. 

11.  Prove  that  the  tetrahedron  formed  by  the  coordinate  planes 
and  a  tangent  plane  to  the  surface  xyz  =  a^  is  of  constant  volume. 

12.  Find  the  angle  at  which  the  normal  to  the  hyperboloid 
y^  —  x^  -{-  4:  z^  =  IQ  at  the  point  (2,  2,  2)  intersects  the  a:^-plane.  Draw 
the  figure. 

13.  Find  the  equations  of  the  projections  on  the  coordinate  planes 
of  the  normal  to  the  cylinder  x  =  y  +  z~  Sit  (2,  1,  1). 

14.  Find  the  equations  of  the  normal  to  the  surface  x-y  -\-  y^-\-  z^  =  S 
at  the  point  (1,  1,  1). 

15.  Show  that  the  sphere  x-  +  y^  +  z"^  =  2  a'^  and  the  hyperbolic 
cylinder  xy  =  a^  are  tangent  to  each  other  at  the  point  (a,  a,  0). 

16.  Determine  a  and  b  so  that  the  ellipsoid  x-+2y^  +  z^  =  7  and 
the  paraboloid  z  =  ax^  +  hy^  may  intersect  at  right  angles  at  (1,  1,  2). 

Ans.    a  =:  3,  &  =  —  1. 

17.  Find  the  angle  between  the  normal  to  the  oblate  spheroid 
3.2  _|_  ^2  ^  2  5;2  —  10  at  (2,  2,  1)  and  the  line  joining  the  origin  to  that 
point.  Ans,    Arccos  |  V3. 


248  CALCULUS 

18.  In  Ex.  17,  find  the  shortest  distance  from  the  origin  to  the 
normal  in  question.  Ans.    ^V6. 

19.  Find  the  angle  at  which  the  line  -  =  ^  =  -  pierces  the  ellipsoid 

2  a;2  +  i  ?/2  +  s2  ^  2.5.  Ans.  67°  48'. 

20.  Prove  that  every  line  through  the  center  of  a  sphere  intersects 
the  sphere  at  right  angles. 

169.  Space  curves.     Two  surfaces 

(1)  ^(x,  y,  z)  =  0,  ^{x,  y,  2)  =  0 

intersect  in  general  in  a  curve  in  space.  The  curve  is  de- 
termined by  the  equations  of  the  two  surfaces  considered 
as  simultaneous. 

Since  there  are  an  infinite  number  of  surfaces  through  a 
given  curve,  and  since  the  equations  of  any  two  of  these 
surfaces  in  general  determine  the  curve,  it  follows  that  the 
equations  of  the  curve  may  be  given  in  an  infinite  number 
of  ways.  A  particularly  simple  way  is  to  give  the  equa- 
tions of  two  of  the  "  projecting  cylinders  "  —  i.e.  the  cylin- 
ders through  the  curve  with  generators  perpendicular  to 
the  coordinate  planes.  Eliminating  y  and  z  in  turn  be- 
tween the  equations  (1),  we  find  two  equations  of  the  form 

(2)  ct>(ix,z}=0,ylr(x,y}  =  0. 

These  equations  represent  cylinders  through  the  curve  (1) 
with  generators  perpendicular  to  the  a:2-plane  and  the 
2;^-plane  respectively. 

We  have  seen  that  the  coordinates  of  a  point  on  a  plane 
curve  are  frequently  given  in  terms  of  a  parameter  t.  The 
same  device  is  often  employed  with  curves  in  space  :  the 
curve  is  given  by  the  three  parametric  equations 

(3)  X  =/(«),  2/ =  KO.^  =  KO- 

By  eliminating  the  parameter  between  two  different  pairs 
of  these  equations,  we  obtain  equations  of  the  form  (2). 

170.  Tangent  line  and  normal  plane  to  a  space  curve. 
The  tangent  to  the  curve  (1)  of  §  169  at  the  point 
P :  (a:Q,  «/q,  Zq)  is  the  intersection  of  the  tangent  planes  to 


FUNCTIONS  OF  SEVERAL  VARIABLES  249 

the  two  surfaces 

Hence  its  equations  can  be  written  down  at  once. 

The  normal  plane  is  the  plane  through  P  perpendicular 
to  the  tangent.  To  find  its  equation,  we  have  only  to 
transform  the  equations  of  the  tangent  line  to  the 
"  symmetric  form  " 

a  h  c 

after  which  the  equation  of  the  normal  plane  can  be 
written  directly. 

Example :  Find  the  equations  of  the  tangent  line  and 
the  normal  plane  to  the  curve 

^2  _^  ^2  +  2;2  =  3, 

at  the  point  (1,  1,  1). 

The  equations  of  the  tangent  are  found  by  formula  (3), 
§  166,  to  be 

a;  +  ?/  +  2  =  3, 
32;  +  ^  —  2^=2. 

To  put  these  equations  in  the  symmetric  form,  let  us 
eliminate  y  and  z  in  turn,  thus  representing  the  line  by 
two  of  its  projecting  planes  : 

22:-32  =  -l, 

5a^  +  3i/  =  8. 
Equating  the  values  of  x  from  these  two  equations,  we 

find 

_ -3j-h8_3^-l 

or 

3~'  -5  ~     2  '  * 

The  equation  of  the  normal  plane  is  therefore  ^ 

3(:r-l)-5(y-l)4-2(2-l)=0. 


250 


CALCULUS 


171.  Direction  cosines  of  the  tangent.  Let  us  draw  a 
secant  (Fig.  86)  through  P :  (xq,  y^,  Zq)  and  a  second  point 
P'  :  (^Xq  +  Ax,  7/q  +  A^,  Zq  +  A^;),  and  denote  by  s  the  length 
of  the  arc  from  a  fixed  point  to  P,  by  As  the  length  of 
the  arc  PP' .     The  direction  cosines  of  the  secant  are 

,       Ax  ot       ^y  I       Az 

cos  a'  =  ,  cos  p  =      ^  ,  cos  7'  =  . 

pp!  ppi  ppi 

those  of  the  tangent  are 

T        Aa;         T       Aa;     As        c?:c 
cosa=    lim  =    iim ■  =  — , 

pp'^qPP'     pp-^o^s    pp>      ds 

_dy 

ds 

dz 

ds 


(1) 


cos/3  = 
cos  7  = 


Hence  the  direction  cosines  of   the    tangent  are  propor- 
tional to  dx,  dy,  dz. 

From  this  fact  we  obtain  at 
once  the  equations  of  the 
tangent  to  the  curve  (3)  of 
§  169.     They  are 

^~  ^o-_  y  ~  Vq  _  ^  ~  ^0 

/'Co)      ^''(^o)      ^'(<o) 
The  equation  of  the  normal 
plane  may  be  written  down 
at  once. 

172.    Length    of     a     space 
curve.     Since 
cos^  a  4-  cos^  p  4-  cos^  7  =  1, 
it  follows  from  the  formulas  (1)  of  the  preceding  article 
that 


169  be- 


Fig.  86 


ds  =  dx  -\-  dy  -\-  dz  . 
Hence  the  length  of  the  arc  of  the  curve  (3-)  of 


FUNCTIONS  OF  SEVERAL  VARIABLES  251 

tween  any  two  points  Pq  :  (^Xq,  y^^  z^)  and  P^ :  (a^j,  y^,  z^  is 

For  the  curve  (2)  of  §  169  this  becomes 


=rV'-(*)v(r 


M-]dx, 


EXERCISES 

1.  Find  the  equations  of  the  projecting  cylinders  of  the  curve 
x^  4-  y^  =  2  a'^,  x^  —  if-  -\r  z^  —  a^;  also  find  the  equations  of  the  tan- 
gent line  and  the  normal  plane  at  the  point  (a,  «,  a). 

Ans.  Normal  plane :  x  —  3/  —  22;  +  2a  =  0. 

2.  Find  the  equations  of  the  tangent  line  and  the  normal  plane  to 
the  circle  x-  +  2/2  _^  ^^  =  9,  ?/  +  2  =  3  at  the  point  (2,  2,  1). 

3.  Find  the  equations  of  the  tangent  line  and  the  normal  plane  to 
the  helix  x  —  a,  cos  0,  y  =  a  sin  6,  z  =  bd  at  the  point  6  =  6q. 

4.  Find  the  length  of  one  turn  of  the  helix  in  Ex.  3. 

5.  Find  the  angle  between  the  curves 

x^  -\-  y^  +  z^  =  3, 
z  =  xy, 
and 

x^  —  y^  -{-  z^  =  1, 

X  +  y  +  z=  'd, 
at  the  point  (1,  1,  1). 

6.  Find  the  circumference  of  the  circle  4:  x'^  +  3y^  -{-  2z-  =  1,  z  =  x. 

7.  Find  the  condition  that  the  surfaces  ^(x,  y,  z)  =0,  ^(a:,  y,  z)  =  0 
intersect  at  right  angles  in  a  point  (xq,  y^.,  z^) '. 

8.  Find  the  centroid  of  the  arc  of  the  curve  x^  =  2  ay,  x^  =  6  a^z 


from  (0,0,0)  to  ^«,  ^,  ^^ 


CHAPTER  XXII 


ENVELOPES.     EVOLUTES 


173.  Envelope  of  a  family  of  plane  curves.     The  equa- 
tion 

(1)  f(x,  y,  a)  =  0, 

where  a  is  arbitrary,  represents  a  family  of  plane  curves  : 

a  is  constant  for  any  one  curve,  but  varies  when  we  pass 

from   one   curve    of    the    family   to    another.      Thus    the 

equation 

represents  all  the  unit  circles  having  their  centers  on  the 
ir-axis  ;   the  equation 

y  =■  X  -\-  k 
represents  the  family  of  straight  lines  making  an  angle  of 
45°  with  OX. 

It   may   happen   that   there    exists    a    curve    to    which 
each    member    of    the    family    (1)    is    tangent.     Such    a 

curve  is  called  the  envelope 
of  the  family.  The  family  of 
circles  mentioned  above  Have 
the  lines  ?/  =  ±  1  as  their  en- 
velope, since  each  of  the  cir- 
cles is  tangent  to  these  lines. 
On  the  other  hand,  the  fam- 
ily of  straight  lines  y  =^x  -\-h  have  no  envelope. 

174.  Determination  of  the  envelope.     Suppose  that  the 
curves 

(1)  fix.  y,  a)  =  0 

252 


Fig.  87 


ENVELOPES.     EVOLUTES  253 

have  an  envelope.  Let  (2:,  y)  be  the  point  of  tangency 
of  the  envelope  with  a  curve  Q  of  the  family;  then  the 
coordinates  x  and  y  are  functions  of  a  alone,  and  they 
satisfy  equation  (1).     Differentiating  (1),  we  find  (§  165) 

^  ^  dx  dy   ^      da 

We  have  not  yet  made  use  of  the  fact  that  the  envelope 
and  the  curve  C  have  a  common  tangent  at  (x^  y). 
The  slope  of  the  tangent  to  0  at  (a;,  y)  is  determined 
by  the  equation   (§  165) 

(3)  fdx  +  ^dy  =  0, 

dx  dy 

and  this  gives  the  slope  of  the  envelope  also.  Combining 
(2)  and  (3),  we  find 

^da  =  0. 
da 

But  since  x  and  y  are  functions  of  a,  a  is  the  independent 
variable  and  we  may  take  da  =^  0.      We  thus  find 

da 

as  a  second  equation,  in  addition  to  (1),  that  is  satisfied  by 
the  coordinates  x^  y.     Hence  the  equations 

fix,  y,  a)  =  0, 


(^) 


da 


taken  together  constitute  'parametric  equations  of  the  en- 
velope. The  equation  in  cartesian  coordinates  can  be 
found  by  eliminating  the  parameter  a. 

In    the    above    discussion    the    existence    of    the    en- 
velope was  assumed.     It  can  be  shown,  conversely,  that 

the  curve  (4)  is  an  envelope,  provided  -^  and  -^  do  not 
both  vanish  along  the  curve. 


254 


CALCULUS 


Example :  Find  the  envelope  of  the  family  of  straight 


a 


lines  7/  =  mx  H ,  where  m  is  the  variable  parameter. 


m 


Differentiating  partially  with  respect  to  m,  we  find 


0  =  X  - 


a 


m 


2' 


or 


m 


a 

=  ±  V- 

^  X 


Substituting  this  value  of  m  in 
the  original  equation,  we  get 

y  =  ±  2Vaa;, 
or 

?/2  =  4  ax. 

This  agrees  with  the  result  of 

analytic     geometr^^     that     the 

straight    line    y  =  mx  + 


a 


is 


m 


Fig.  88 


tangent  to  the  parabola  y'^  =  \ax 
for  all  values  of  m, 

175.  Envelope  of  tangents. 
Every  curve  may  be  considered  as  the  envelope  of  its 
tangents,  as  appears  at  once  from  the  definition  of  the 
envelope.  This  is  illustrated  by  the  example  of  the  pre- 
vious article,  where  the  parabola  was  found  as  the  enve- 
lope of  its  tangents. 

EXERCISES 

Find  the  envelope  of  each  of  the  following  families  of  curves.     In 
each  case  draw  several  curves  of  the  family,  and  the  envelope. 

1.  The  circles  of  radius  a  with  their  centers  on  the  ^-axis. 

2.  The  family  of  straight  lines  y  =  1  mx  +  r»/*. 

Afis.   16  3/3  +  27^:4  =  0. 

3.  The  family  of  parabolas  3/^  -  a{x  -  a).  Ans.  x  ±2  y  =  0. 

4.  The  family  of  circles  whose  diameters  are  double  ordinates  of 
a  parabola. 


ENVELOPES.     EVOLUTES  255 

5.  The  family  of  circles  tangent  to  the  a:-axis  and  having  their 
centers  on  the  parabola  y  =  x^.  Ans.  y  =  0,  2  x^  +  2  y^  -  y  z=z  0. 

6.  The  circles  with  centers  on  a  parabola  and  passing  through  the 
vertex  of  the  parabola.  Ans.  y%x  -\-  2  a)+  x^  =  0. 

7.  The   circles    through   the   origin   with    their   centers   on    the 
hyperbola  x^  -  y^  =  a^  A7is.  (x2  +  ?fy=  4  a\x^  -  y'). 

8.  The  family  of  ellipses  whose  axes  coincide  and  whose  area  is 
constant.  Ans.  Two  conjugate  rectangular  hyperbolas. 

9.  A  straight  line  segment  of  constant  length  moves  with  its  ends 
in  two  perpendicular  straight  lines.     Find  its  envelope. 

Ans.  A  hypocycloid  of  four  cusps. 

10.  A  straight  line  moves  so  that  the  sum  of  its  intercepts  on  the 

111 
axes  is  constant.     Find  its  envelope.     Ans.  The  parabola  a:^  +  y^  —  ^2^ 

11.  When  a  projectile  is  fired  from  a  gun  with  an  initial  velocity 

Vq  inclined  at  an  angle  a  to  the  horizontal,  the  equation  of  its  path,  all 

resistances  being  neglected,  is  (§  235) 

px 
y  —  X  tan  a  —        ^ 


2  vq^  cos^  a 

Find  the  envelope  of  all  possible  trajectories  when  the  angle  of  ele- 
vation a  varies.  Ans.  The  parabola  y  =  — ^ —  :f— - . 

2g     2  v^^ 

12.  The  sides  of  a  variable  right  triangle  lie  along  two  fixed  lines. 
If  the  area  of  the  triangle  is  constant,  find  the  envelope  of  the 
hypotenuse. 

13.  Find  the  equation  of  the  curve  tangent  to  the  lines 

y  =  mx  —  am^, 
where  m  is  the  parameter. 

14.  Find  the  equation  of  the  curve  which  is  tangent  to  the  line 

yoy  =  2ax  -\-  ^  y^^ 
for  all  values  of  yo. 

15.  Find  the  equation  of  the  curve  tangent  to  the  family  of  straight 
lines 

X  cos  ct  -{-  y  sin  cc  =  p, 
where  «  is  the  variable  parameter. 

16.  Find  the  equation  of  the  curve  tangent  to  the  straight  line 


y  =  mx  ±  ^ ahr^  +  IP- 
for  all  values  of  m.  Ans.  —-1-^  =  1. 


256 


CALCULUS 


176.  The  evolute.  When  a  point  P  moves  along  a 
curve,  the  center  of  curvature  Q  (§  54)  describes  a  second 
curve,  called  the  evolute  of  the  original  curve. 

It  can  be  shown  that  the  normal  PQ  to  the  original 
curve  is  tangent  to  the  evolute;  i.e.  the  evolute  is  the 
envelope  of  the  normals.  Its  equation  may  therefore  be 
found  by  writing  the  equation  of  the  normal  to  the  given 
curve  in  terms  of  a  parameter,  and  then  applying  the 
method  of  §  174. 

Example  :  Find  the  evolute  of  the  parabola  «/^  =  4  ax. 

It  is  shown  in  analytic  geometry  that  the  line 
(1)  y  =  mx  —  2  am  —  am^ 

is  normal  to  this  parabola  for  all  values  of  m.  We  have 
therefore  to  find  the  envelope  of  the  family  (1),  regarding 
m  as  the  variable  parameter. 

Differentiating  partially  with  respect  to  m,  we  get 


0  =  X  —  2  a  —  S  am\ 


or 


m^  = 


x  —2  a 
3a 


Fig.  89 


Equation  (1)  may  be  written 
in  the  form 

y  =  m(x  —2  a—  am^)^ 
or 

^2  —  rri^Qx  —2  a—  am^}^. 
^      Substituting  for  m^,  we  find 
the  equation  of  the  evolute 

a    "  semi-cubical    parabola  " 
with  a  cusp  at  (2  a,  0). 


ENVELOPES.     EVOLUTES  257 


EXERCISES 

1.  Find  the  equation  of  the  evolute  of  the  parabola  y^  =  4:ax  by 
writing  the  equation  of  the  normal  in  terms  of  the  ordinate  of  the 
point  at  which  the  normal  meets  the  curve. 

2.  In  Ex.  1,  show  that  the  distance  from  any  point  P  on  the 
parabola  to  the  corresponding  point  on  the  evolute  is  equal  to  the 
radius  of  curvature,  thus  verifying  that  the  locus  of  the  center  of 
curvature  and  the  envelope  of  the  normals  are  the  same  curve. 

x^       iP" 

3.  Find  the  evolute  of  the  ellipse  "^  +  7^  =  1»  given  that  the  equa- 
tion of  the  normal  in  terms  of  the  eccentric  angle  <^  is 

hy  =  ax  tan  <ji  —  (a^  —  b^)  sin  <f>. 

Ans.    (ax)^  +  (byy  =  (a^  -  &2)f. 

2  2  1 

4.  Find  the  evolute  of  the  hypocycloid  x'^  +  y^  =  a^,  the  equation 
of  whose  normal  is 

y  cos  a  —  X  sin  a  =  a  cos  2  «. 

2.  2.  2. 

Ans.    (x  +  y)^  +  {x  —  y)^  =  2  a^. 


CHAPTER   XXIII 


MULTIPLE   INTEGRALS 


177.    Volume  under  a  surface.     Let  us  try  to  find  the 
volume  V  bounded  by  a  portion  T  of  the  surface 

the  area  S  into  which  T  projects  in  the  2:?/-plane,  and  the 
cylindrical  surface  through  the  boundaries  of  S  and  T. 

Z 


Fig.  90 


We  can  get  an  approximate  expression  for  the  required 
volume  as  follows.     Draw  in  S  a  set  of  n  lines  parallel  to 

258 


MULTIPLE  INTEGRALS  .259 

the  «/-axis  and  a  set  of  m  lines  parallel  to  the  a;~axis,  as 
in  Fig.  90,  thus  dividing  S  into  rectangles  of  area  A«/  A2;, 
together  with  a  number  of  irregular  portions  around  the 
boundary.  By  passing  through  each  line  of  the  two  sets 
a  plane  perpendicular  to  the  a;?/-plane  we  divide  V  into 
vertical  rectangular  columns,  together  with  smaller  ir- 
regular columns.  The  upper  boundary  of  each  rectangu- 
lar column  is  a  portion  of  the  surface  T.  Through  that 
point  of  the  upper  boundary  of  each  column  which  is 
nearest  the  :ri/-plane,  pass  a  horizontal  plane,  thus  form- 
ing a  set  of  rectangular  prisms  lying  wholly  within  V. 
The  sum  of  the  volumes  of  these  prisms  is  evidently 
an  approximation  to  the  required  volume,  the  error  com- 
mitted being  the  sum  of  the  irregular  columns  around 
the  outside,  together  with  the  portions  lying  above  the 
upper  bases  of  the  rectangular  prisms.  That  is,  approxi- 
mately^ 

where /(a;<,  ^,)  is  the  altitude  of  the  prism. 

It  is  obvious  that  the  error  in  this  approximation  may 
be  made  arbitrarily  small  by  taking  both  b^x  and  A?/  suf- 
ficiently small.     Hence  the  required  volume  is  exactly 


n      m 


(1)  r=  lim  V  VKx,,  y^l^y  Ax. 

The  "double  limit"  (1)  may  be  evaluated  by  two 
successive  applications  of  the  fundamental  theorem  of 
§  104,  as  follows. 

Let  us  fix  our  attention  on  the  rectangle  P-Pl'Ql'QI  in 
S  (Fig.  90).  The  volume  AF/  whose  base  is  this  rec- 
tangle may  be  found  approximately  by  adding  the  volumes 
of  all  the  included  elementary  prisms.  Hence,  by  the 
fundamental   theorem    of    §  104,    AF"/   is   given    exactly 


260.  CALCULUS 

by  the  formula 

m 

AF/  =  lim   ^f(Xi,  y>}^yi^x 

Xi  and  ^x  remaining  constant  as  we  pass  to  the  limit. 

Now  if  we  add  all  the  volumes  of  this  type,  we  have 
approximately  the  required  volume.  It  is  to  be  noticed 
that  in  the  expression  for  A  Vl  the  coefficient  of  ^x  is  a 
function  of  Xi  alone,  since  the  limits  yl  and  yl'  are  func- 
tions of  Xi  alone.  Thus  we  may  apply  again  the  theorem 
of  §  104,  and  find  that  the  required  volume  under  the  sur- 
face z  =f(x,  y)  is 


dx^ 


F=  lim  5     P  f{Xi,  y)dy   Ax  =  f  \   P  f(x,  y)dy 

where  a  and  h  are  the  extreme  values  of  x  on  the  bound- 
ary of  S. 

The  quantity  just  found  is  usually  written  without  the 
brackets,  thus : 

(2)  V  =  j     I      f(ix,y)dydx. 

It  is  called  a  double  integral^  or  more  properly  an  iterated 
integral^  being  merely  an  integral  of  an  integral.  It 
is  to  be  noted  that  the  inner  integral  sign  belongs  with 
the  inner  differential,  and  that  during  the  integration 
with  respect  to  ?/,  x  remains  constant.  Further,  the 
first  or  inner  limits  of  integration  are  in  general  variables, 
but  the  outer  limits  are  always  constants. 

Of  course  we  might  integrate  first  with  respect  to  x^ 
then  with  respect  to  y.  The  same  argument  as  before 
would  lead  to  the  formula 

(3)  V=   rr"Ax,y)dxdy, 

y  remaining  constant  during  the  first  integration. 


MULTIPLE  INTEGRALS 


261 


In  the  foregoing  argument,  we  have  assumed  our  vol- 
ume to  be  divided  into  rectangular  columns  perpendicular 
to  the  a^?/-plane.  Frequently,  however,  it  is  more  conven- 
ient to  erect  columns  perpendicular  to  one  of  the  other 
coordinate  planes  (cf.  example  (5)  below).  Such  varia- 
tions offer  no  difficulty  provided  the  geometric  meaning 
of  the  successive  integrations  be  kept  clearly  in  mind.  In 
every  problem,  a  sketch  of  the  required  volume  should  be 
made,  and  the  required  double  integral  built  up  by  inspec- 
tion of  the  figure. 

Examples:    («)    Find  the   volume  in   the   first   octant 
bounded  by  the  plane  z  =  x-\-  y 
and  the  cylinder  y  =  1  —  x'^. 

Integrating  in  the  order  ?/,  a;, 
we  have 


•l-J-2 


=  i    i        (x  +  y)dydx 


=Xi^^+% 


r 


1— X2 


dx 


-£{-^-'-^^> 


—  31 
~  60* 


Fig.  91 


(5)  Find  the  volume  common  to  the  circular  cylinder 
?/2  _|_  ^2  —  ay  and  the  sphere  x"^  -{-  y^  -\-  z^  =  a^. 

Let  us  divide  the  volume  into  columns  perpendicular  to 
the  ^z-plane  : 

X  dzdy 


Va*^  —  y'^  -~  z^  dz  dy^  etc. 


262  CALCULUS 

178.  Volume  under  a  surface  :  second  method.  The  re- 
sult of  §  177  may  be  obtained  by  a  somewhat  different 
method.  The  area  of  the  section  by  a  particular  one  of 
the  planes  parallel  to  the  i/z-plaue  is  evidently 

Hence,  by  §  110,  the  volume  is 

A(x)dx=  I     I     f(x^y)dydx,   . 

a  %/a*/y' 

The  actual  work  of  obtaining  the  volume  in  any  particular 
case  is  therefore  the  same  by  the  two  methods  —  the  only 
difference  is  in  the  geometric  interpretation  of  the  succes- 
sive steps.  The  great  advantage  of  the  method  of  §  177 
is  that  it  lends  itself  readily  to  the  discussion  of  a  great 
variety  of  other  problems  besides  the  computing  of  vol- 
umes, as  we  shall  see  in  the  next  few  articles. 

Of  course  when  ^(rr)  is  known  to  start  with,  the  volume 
may  be  found  by  a  single  integration  as  in  §  110.  This  is 
the  case  in  several  of  the  exercises  below. 

EXERCISES 

In  each  of  the  following  exercises,  the  limits  of  integration  should 
be  obtained  directly  from  a  figure. 

1.  Find  the  volume  in  the  first  octant  bounded  by  the  planes 
x  =  l,  z  =  x-\-y  and  the  cylinder  y'^  =  x.  Ans.  y . 

2.  Find  the  volume  in  the  first  octant  bounded  by  the  cylinder 
a:^  +  ?/2  =  a^  and  the  plane  z  =  x  +  y.  Ans.  f  a^. 

3.  Find  the  volume  of  a  cylindrical  column  standing  on  the  area 
common  to  the  two  parabolas  x  =  y'^,  y  =  x'^  as  base  and  cut  off  by  the 
surface  2  =  1  +  ?/  —  x^.  Check  the  result  by  integrating  in  two  ways  : 
first  in  the  order  y,  x ;  next  in  the  order  x,  y. 

4.  Find  the  volume  in  the  first  octant  bounded  by  the  plane 
y  +  z  =  1  and  the  surface  x  =  i  —  z  —  y^.     Check  as  in  Ex.  3. 

5.  Find  the  volume  cut  off  from  the  paraboloid  y  =  1  ~  y  ~  77 
by  the  xz-^lane. 


MULTIPLE  INTEGRALS  263 

6.  Find  the  volume  in  the  first  octant  under  the  surface  z  —  xy 
bounded  by  the  cylinder  ij  —  x^  and  the  plane  ?/  =  1.  Solve  in  two 
ways, 

7.  Find  the  volume  bounded  by  the  surface  z  =  xy,  the  cylinder 
2/2  =  ax,  and  tlie  planes  x-\-y  =  2a,  y  —  0,  z  =  0. 

8.  Find  the  volume  sliced  off  from  the  paraboloid  az  =  a^  —  x^  —  y^ 
by  the  plane  y  -\-  z  =  a. 

9.  Find  the  volume  cut  out  of  the  first  octant  by  the  cylinders 
z  —  \  —  x'^,  X  =  \  —  y'^.  Ans.  \%. 

10.  Find    the    volume   of.  a    segment    of    an    elliptic    paraboloid 
bounded  by  a  right  section. 

11.  Find    the    volume    bounded   by   the  surfaces  4  y'^  -\-  4:  z^  =  x^, 
X  =  4:  y,  X  =  2  a,  z  =  0.  Ans.   sVC-Itt  —  3V3)a^. 

12.  Find  in  two  ways  the  volume  in  the  first  octant  bounded  by 
the  paraboloid  y  =  xz  and  the  planes  z  =  x,  z=  2  —  x. 

13.  Find  the  volume  of  a  segment  of  a  hyperboloid  of  two  nappes 
bounded  by  a  right  section. 

2  2^2 

14.  Find  the  volume  bounded  by  the  surface  x^  -\-  y^  -{■  z^  =  a^. 
Sketch  this  surface.  Ans.  /^  tt  a^. 

15.  Find   the    volume    in    the    first    octant   inside   the   cylinder 
x^  +  y''^  =  2  ax  and  outside  the  paraboloid  x^  +  y'^  =  az.        A71S.   |  tt  a^. 

16.  Find  the  volume  in  the  first  octant  bounded  by  the  surface 
111 


a  J        \hj         \cJ  90 

17.  Show  that  the  volume  of  any  cone  or  pyramid  is  one  third  the 
area  of  the  base  times  the  altitude. 

18.  Write  out  six  different  double  integrals  for  the  volume  in  the 
first  octant  bounded  by  the  cylinders  y  =  x-,  x^  -{■  z'^  =  1. 

19.  Find  the  volume  in  Ex.  18  by  simple  integration. 

179.  Interpretation  of  the  given  function.  Any  function 
/(a;,  z/)  of  two  independent  variables  may  be  interpreted 
as  the  2-coordinate  of  a  point  on  a  surface  in  space. 
If,  then,  in  any  problem,  we  can  express  the  required 
quantity  as  -a  double  limit  of  tlie  form  (1),  §  177,  no 
matter  ivliat  may  he  the  geometric  or  physical  meaning  of  the 
given  function  f(x^  ?/),  the  limit  may  be  evaluated  by  an 


264 


CALCULUS 


iterated  integration  as  in  §  177.  Thus  the  result  of  that 
article  is  by  no  means  confined  to  the  determination  of 
volumes  —  we  shall,  as  was  mentioned  in  §  178,  apply  it 
to  the  study  of  a  variety  of  problems. 

180.    The  double  integraL     In  the  argument  of  §  177  it 
is  not  necessary  that  the  function  /  be  expressed  in  terms 

of  cartesian  coordinates  x  and 
y ;  further,  the  area  S  need  not 
be  divided  into  elements  in  the 
particular  way  there  adopted. 
The  essential  points  are,  first, 
that  we  have  a  function  /  of 
two  independent  variables  de- 
fined at  all  points  of  the  region 
S\  second,  that  we  divide  S 
into  n  elements  AaS'  which  are 
infinitesimal  of  the  second 
order.^ 

When  S  is  divided   in  this 
way,  the  double  limit 


Fig.  92 


lim  V/,A^ 


is   called   the  double  integral  of  the   function  /  over   the 

region  aS',  and  is  denoted  by  the  symbol  j  j  f  dS : 

s 

As  noted  in  §  177,  the  integral     I     I  ^   f(x^  V^^y  dx  is 

often  called  a  double  integral,  and  it  is  evidently  equal  to 

j   \  f  dS  \  but  it  is  clear  that  the  latter  integral  is  the 


*  That  is,  such  that  the  maximum  distance  between  two  points  on  the 
boundary  of  A  6^  approaches  0. 


MULTIPLE  INTEGRALS 


265 


more  general,  since  it  does  not  tie  us  down  to  a  particular 
coordinate  system,  or  to  a  particular  mode  of  division  of  S. 
The  integrals  (2)  and  (3)  of  §  177  are  merely  two  special 
forms  of  the  double  integral. 

181.    The  double  integral  in  polar  coordinates.     Given  a 
function  /(>,  6)  of  the  polar  coordinates  r,  ^,  the  double 

integral    I   (  /  dS  may  be  evaluated  as  follows.     Divide  jS 

s 
into  elements  by  a  set  of  circles  with  center  at  the  origin 
and  a  set  of  lines  radiating  from 
the  origin,  as  in  Fig.  93.     Then 
AS  is  the  difference  between 
two  circular  sectors  of  angle  A^ 
and  radius  r  and  r  +  Ar  respec- 
tively ;  i.e. 
AS  =  l(r  +  Ar)2A6'  -  ~i-  r^AO 
=  (rAr  4-  J  aP)AO. 

We  may  now  repeat  the  argu- 
ment of  §  177,  integrating  first 
with  respect  to  r,  and  noting 
that,  by  §  109,  the  infinitesimal 

of  higher  order  ^aPaO  may  be  neglected.  This  leads  to 
the  result 

fffdS=lim^^f(r,  e)rArAe  =f^X^'fir,  Q}rdrdB. 

s 

EXERCISES 


Fig.  93 


1.  A  round  hole  is  bored  through  the  center  of  a  sphere.  Find 
the  volume  cut  out,  using  polar  coordinates. 

2.  A  cyhnder  is  erected  on  the  circle  ?-  =  a  cos  ^  as  a  base.  Find 
the  volume  of  the  cylinder  inside  a  sphere  of  radius  a  with  center  at 
the  origin. 

3.  Find  the  volume  above  the  a;y-plane  common  to  the  paraboloid 
2  =  4  —  x^  —  r/2  and  the  cylinder  x^  ■}-  ij-  =  1,  using  polar  coordinates. 


266  CALCULUS 

4.  A  square  hole  of  side  2  whose  axis  is  the  2-axis  is  cut  through 
the  paraboloid  of  Ex.  8.     Find  the  volume  cut  out. 

5.  Find  the  volume  of  a  spherical  wedge  by  double  integTation. 

6.  Prove  that  when  a  curve  r  =f(0)  revolves  about  the  initial 
line,  the  volume  of  revolution  generated  is  given  by  the  formula 

F  =  2  TT  P  f '   r  sin  6-rdr  d6. 

Solve  the  following  by  the  method  of  Ex.  6. 

7.  Find  the  volume  of  a  sphere. 

8.  Find    the     volume     generated    by     revolving     the     cardioid 
r  =a(l  —  sin  ^)  about  its  line  of  symmetry,  Ans.  |7ra^. 

9,.    The  curve  r^  =  a^  sin  6  revolves  about  the  y-axis.     Find   the 
volume  generated. 

10.  Find  the  volume  generated  by  revolving  one  loop  of  the  curve 
r  =  a  cos  2  $  about  its  line  of  symmetry. 

11.  Find  the  volume  generated  by  revolving  a  circle  about  one  of 
its  tangents. 

12.  Find  the  volume  cut  from  a  sphere  by  a  cone  of  half-angle  ^ 

o 

with  its  vertex  at  the  center  of  the  sphere.     Check  by  using  cartesian 
coordinates. 

13.  Find  the  volume  of  the  prolate  spheroid  generated  by  revolv- 
ing about  its  major  axis  the  ellipse 

I 


r  — 


1  —  e  cos  0 


A       7S 

where  e  is  the  eccentricity.  Ans.   — 

3(l_e2)2 

14.  Find  the  volume  of  a  paraboloid  of  revolution  bounded  by  a 
right  section  through  the  focus,  taking  the  equation  of  the  generating 

0  n 

parabola  in  the  form  r  =  — -•  Ans.  2  7ra^. 

1  —  cos  6 

182.  Transformation  of  double  integrals.  We  have  seen 
that  the  integrals  (2)  and  (3)  of  §  177  are  merely  dif- 
ferent  forms   of   the    double    integral    j   I  fdS.     It  may 

s 
happen  that  an  integral  given  in  the  form  (2)  is  difficult 

or  impossible  to  evaluate,  but  that  when  transformed  to 
the  form  (3),  it  becomes  simple.  Or  sometimes  after 
evaluating  the  form  (2)  we  change  to  the  form  (3)  and 


MULTIPLE  INTEGRALS 


267 


evaluate  again,  merely  as  a  check  on  the  result.  The 
process  of  changing  the  form  of  an  integral  from  (2)  to 
(3),  or  vice  versa^  is  called  inverting  the  order  of  integration. 
Another  transformation  of  importance  is  the  change 
from  one  coordinate  system  to  another  —  for  instance, 
from  cartesian  to  polar  coordinates. 

Example  :   Evaluate    I     \    —  cIt/  dx. 

This  integral  cannot  be  evaluated  di- 
rectly,  since  the  function  —  is  not  inte- 

y 

grable  in  terms  of  elementary  functions. 
But  a  study  of  the  limits  shows  that 
the  field  of  integration  is  the  triangle 
bounded  by  the  lines  2:  =  0,  y  —  x^  y  =  a. 
Hence 


Fig.  94 


I     I     —ay  dx 

c/Q   *Jx      y 


*^o  •^o    y 


■s 

Jo 


eydy  =  e°  —  1. 


y 


X 


dy 


EXERCISES 

1.  Check  the  result  in  example  (a),  §  177,  by  inverting  the  order 
of  integration. 

2.  Invert  the  order  of  integration  in  Exs.  1  and  2,  p.  262. 

3.  Find   the   volume  bounded  by  the  cylinder  x'^  =  4  ay  and  the 
planes  x  +  y  +  z  =  a,  z  =  0,  a:  =  0,  integrating  in  two  different  ways. 

4.  Express  the  volume  of  Ex.  3  as  a  double  integral  in  two  other 
ways. 

5.  Interpret   the    integral    \  "  i      "  V4  a^  —  y^  dy  dx  as  a  volume, 

and  write  out  five  other   double  integrals   (all  in  cartesian    coordi- 
nates) for  this  same  volume. 

6.  Evaluate    i  "  (  "  ^(.^  +  y^dy  dx,    and   check   by  inverting   the 
order  of  integration.     Interpret  geometrically. 

7.  Evaluate   C  ^^ e'''' dy  dx.  Ans.  0.859. 


268 


CALCULUS 


8.   Evaluate   \     \ 

Jo  Ji 

dmates. 


1  /'V^i-a 
'O 


9.   Evaluate 


sin 
1  r2         2 


6^2+2/2  (ffj  fix  by  transforming  to  polar  coor- 

Ans.  1.35. 

TTiJ 


Jo  J-z 


r 


■  dy  dx. 


,         4 
Ans.    —• 


ir 


10.  Express   \  i/(/5  (a)  in  cartesian  coordinates, (&)  in  polar  coor- 

s 
dinates,  where  >S'  is   the   triangle  bounded  by  tlie  lines  x  =  a,  y  =  0, 
y  =  x. 

/•  2     /•  o  cos  S 

11.  Transform    4  "^  I       e''*^"<>  r^/r//^  to  cartesian  coordinates. 

Jo    Jo 

12.  Compute  the  value  of   I   |  cos  (x^  +  y'^)dS  extended  over  the 

interior  of  the  circle  x^  +  ?/'-=  1,  •  Ans.  2.644. 

13.  Find  the  area  in  the  first  quadrant  under  the  curve  y  =  e'2'^  by 
noting  that 


'^^'dxyU^e-'-y'dyy^^^^Je-'-^^^y'-\lydx 

Ans.  ^^ 


Fig.  95 


14.  Find  the  centroid  of 
the  area  in  Ex.  13. 

183.  Area  of  a  sur- 
face. Let  us  try  to  find 
the  area  cr  of  a  portion 
of  the  surface 

Y    Suppose  that  the  xy-^vo- 

jection  of  a  is  the  region 
aS'.  Let  us  divide  S  into 
elements  AaS'  in  any  suit- 
able way,  and  fix  our 
attention  on  a  particular 
one  of  these  elements. 
This  element  is  the  hori- 
zontal projection  of  tlie 


MULTIPLE  INTEGRALS  269 

portion  Act  of  a.  If  we  draw  the  tangent  plane  at  some 
point  P  of  Act,  then  AaS'  will  be  the  horizontal  projection 
of  a  certain  area  Act'  on  the  tangent  plane.     Hence 

AaS'  =  Ao-'  cos  7, 

where  7  is  the  angle  between  the  2-axis  and  the  normal 
PiV^to  o-  at  P,  and 

Ao-'  = 

cos  7 

If  now  we  form  the  sum  of  the  quantities  Act'  and  pass 
to  the  limit,  we  have 

^s^o^'^cosY     ^^-^  cos  7 

In  case  it  is  more  convenient  to  project  the  area  on  the 
xz-  or  the  ^5;-plane,  the  corresponding  formula  is  readily 
developed. 

Example :  Find  the  area  of  that  part  of  the  surface 
z=  y  -\-x'^  whose  projection  on  the  :c?/-j)lane  is  the  triangle 
bounded  by  the  lines  y  =  ()^  y  —  x,  x  =  \. 

Writing  the  equation  of  the  surface  in  the  form 

2  —  ^  —  2:^  =  0, 
we  have  for  the  partial  derivatives  the  values 

dF         o        ^F         .      OF     . 
—  =  —  zx^    —  =  — 1,    — =1- 
dx  dy  dz 

Hence  by  the  theorem  of  §  167,  the  direction  cosines  of 
the  normal  are  proportional  to  —  2  2:,  —  1 ,  1,  and 

1 

cos  7  =  —  • 

V4  2:2  4-  1  +  1 

Therefore  the  required  area  is 

o-  =f^f^  V4  x^  +  2  dy  dx 

=   f  xV4:  x^  H-  2  dx 

Jo 


=  l.f(4a-2+2)- 


=  ^5(6^- 20- 


270 


CALCULUS 


EXERCISES 

1.  Find   the    area  cut  out  of   the  plane  x-\-y  +  2z  =  2a  by  the 
cylinder  a;2 +  ?/-  =  rt-.  Ans.    ^VQ-n-a'^. 

3 

2.  Find  the  area  of  that  part  of  the  surface  z  =  y  +  ^x^  whosfe 
projection  in  the  ^-^/-plane  is  the  triangle  bounded  by  the  lines  y  =  0, 

Ans.   if (2  +\/2). 

The  center  of  a  sphere  of  radius  a  is  on  the  surface  of  a  cylinder 


y  =  X,  X  =  2 


Find  the  surface  of  the  cylinder  intercepted  by  the 

Ans.    4  rt^. 


of  diameter  a. 
sphere. 

4.  In  Ex.  3,  find   the    surface  of   the  sphere   intercepted  by  the 
cylinder.  Ans.    2(7r  —  2)a^. 

5.  How  much  of   the   conical   surface   z^  =  x^  -\-  y^   lies    above   a 
square  of  side  2  a.  in  the  x?/-plaue  whose  center  is  the  origin  ? 

6.  How    much   of   the   surface    az  =  xy   lies  within    the  cylinder 
3.2  ^  y2  —  f^2  9     (Use  polar  coordinates.) 

7.  A  square  hole  is  cut  through  a  sphere,  the  axis  of  the  hole  coin- 
ciding with  a  diameter  of  the  sphere.     Find  the  area  cut  from  the 


surface  of  the  sphere. 


Ans.    16rt&arcsin 


8  a^  arcsin 


&2 


62 


184.    Triple  integrals. 
z 


Fig.  9G 


We  have  seen  that  the  integral 
of  a  function  of  one 
variable,  extended 
over  a  given  inter- 
val, may  be  inter- 
preted as  the  area 
under  a  plane 
curve.  Again,  the 
integral  of  a  func- 
tion of  two  varia- 
bles extended  over 
a  plane  region  may 
be  interpreted  as 
the  volume  under 
a  surface.  If  now 
we  have  a  function 
of  three   variables 


MULTIPLE  INTEGRALS  271 

defined  at  all  points  of  a  portion  of  space,  no  similar  geomet- 
ric interpretation  for  the  integral  of  the  function  over  the 
given  region  is  possible,  since  geometric  intuition  fails  in 
space  of  four  dimensions.  Nevertheless  the  meaning  of 
such  an  integral  may  be  made  plain  by  analogy  with  the 
earlier  cases. 

Suppose  we  have  given  a  function  /(a:,  ?/,  z)  defined  at 
all  points  of  a  three-dimensional  region  V.  Let  us  pass 
through  V  three  sets  of  planes  parallel  to  the  coordinate 
planes,  thus  dividing  V  into  elementary  rectangular 
parallelepipeds  of  volume  Aa:  A?/  A2,  together  with  smaller 
irregular  portions  around  the  boundary.  Now  multiply 
the  volume  of  each  element  by  the  value  of  the  function 
at  some  point  within  the  element,  say  at  its  center,  and 
form  the  sum  of  these  products.     The  triple  limit 

Aa;->0  ^^  ^^  ^^ 

is  defined  as  the  value  of  the  trijjle  integral  of  /(a:,  ?/,  z) 
over  the  region  V. 

This  limit  may  be  evaluated  by  three  successive 
integrations  (cf.  §177)  : 

r  =  lim  y  V  y  fix.  y,  z)Ax  Ay  Az 

Ax->0  ^^  "^  '^ 


I        I     fix,y,z)dzdydx. 

a  %Jy     ^ z' 


The  first  integration  extends  over  a  vertical  column  of 
base  Ay  Ax\  the  limits  z\  z"  are  the  extreme  values  of  z 
in  this  column,  and  are  in  general  functions  of  both  x  and 
y.  The  integration  with  respect  to  y  is  extended  over 
a  slice  parallel  to  the  ?/2-plane ;  the  limits  y'  and  ^"  are 
the  extreme  values  of  y  in  this  slice,  and  are  functions  of 
X  alone.  In  the  final  integration  the  limits  are  of  course 
the  extreme  values  of  x  in  the  whole  region. 


272  CALCULUS 

More  generally,  the  function  /  may  be  given  in  terms 
of  any  system  of  coordinates,  and  the  region  V  may 
be  divided  into  elements  in  any  suitable  way.*  We 
write  in  general,  for  the  value  of  the  triple  integral  of  / 
over  the  region  F^ 

V 

It  is  hardly  necessary  to  say  that  such  transformations 
as  inversion  of  order  and  change  from  one  coordinate 
system  to  another  are  allowable  and  useful  with  triple 
integrals,  just  as  with  double  integrals. 

It  may  be  well  to  observe  at  this  point  that  applica- 
tions of  triple  integration  are  comparatively  rare  in  ele- 
mentary work.  In  the  problems  treated  in  the  next  two 
articles,  triple  integrals  are  sometimes  required. 

The  volume  1^  itself  may  be  expressed  as  a  triple  inte- 
gral, the  given  function/ being  taken  equal  to  unity  : 

V 

It  is  true  that  the  volume  may  be  found  more  directly  by 
methods  previously  studied  ;  nevertheless  it  may  be  worth 
while  to  solve  a  few  exercises  by  the  present  method  for 
the  sake  of  practice  in  determining  the  limits  in  triple 
integration. 

Example :  Find  the  volume  cut  off  from  the  paraboloid 

z  =  l  —  x^  —  ^ 
by  the  a;j/-plane.  ^ 

In  this  case 


*  The  element  must  of  course  be  infinitesimal  of  the  third  order. 


MULTIPLE  INTEGRALS  273 


3*8*2 


3' 

12 


2\/l-x2  , 

ax 


EXERCISES 

Finci  the  following  volumes  by  triple  integration,  drawing  a  figure 
in  each  case. 

1.  The  tetrahedron  bounded   by  the   coordinate   planes  and   the 

,         X      y      z      ^ 
plane  -■{■•}  +  -=  I. 
a      0      c 

2.  The  volume  bounded  by  the  paraboloid  x^  -\-  y"^  =  az,  the  cylin- 
der a:-  +  y-  =  2  ax,  and  the  plane  2  =  0.  '  Ans.   ^  na^. 

3.  Interpret  the  triple  integral  \  \  \  ^'~^  dz  dy  dx  geometri- 
cally, and  express  the  same  volume  as  a  triple  integral  in  several  other 
ways,  drawing  a  figure  for  each  case. 

185.  Heterogeneous  masses.  The  density  of  a  homo- 
geneous mass  has  been  defined  in  §  121  as  the  ratio  of 
the  mass  to  the  volume  it  occupies : 

V 

For  a,  heterogeneous  mass,  i.e.  one  whose  density  varies 
from  point  to  point,  we  must  introduce  the  idea  of  density 
at  a  point. 

Consider  an  element  of  volume  A  F^  including  a  point  P, 
and  let  Ail[f  denote  the  mass  contained  in  AF".     Then  the 

ratio  — -  is  the  average  density  in  A  Fi     If  A I^^ approaches 

0  in  such  a  way  that  P  is  always  included,  the  ratio  

in  general  approaches  a  limit  *  S,  called  the  density  at  the 

*  In  general  this  is  true  only  if  AT" is  infinitesimal  of  the  third  order, 
as  in  §  184. 

T 


274  CALCULUS 

point  P : 

5  ,.      AM     dM 

6  =   lim  - —  = 

AF->oA7     dV 

The  mass  of  a  heterogeneous  body  whose  density  at  any 
point  is  given  as  a  function  of  the  coordinates  of  the  point 
can  be  found  by  integration.  We  have  only  to  choose  a 
suitable  mass-element  and  integrate  over  the  whole  body. 
The  great  point  to  be  noted  is  that  in  general  the  element 
itself  must  be  homogeneous^*  since  otherwise  the  mass  of 
the  element  cannot  be  computed  and  hence  the  integral 
cannot  be  built  up. 

In  many  cases  it  is  possible  to  choose  an  element  as  in 
Chapter  XV  and  obtain  the  result  by  a  simple  integra- 
tion ;  in  more  complicated  problems  double  or  triple  in- 
tegration may  be  necessary. 

We  give  the  argument  in  full  only  for  the  general  case 
where  triple  integrals  are  employed.  Given  a  mass  M 
occupying  a  volume  FI  divide  I^into  elements  as  in  §  184, 
and  multiply  each  element  A  J^  by  the  density  B  at  one  of 
its  points.     Then  the  sum  ^^^^SAFis  an  arbitrarily 

close  approximation  to  the  mass  Mii  AP^be  taken  suffi- 
ciently small,  and  the  mass  is  therefore  given  exactly  by 
the  formula 

r 

For  a  mass  distributed  over  a  surface  S,  the  idea  of 
"surface  density"  must  be  introduced: 
g^  lim  Aiff  ^  dM 

whence,  by  argument  now  familiar, 

M  =  ffBdS. 

*  By  this  is  meant  that  the  density  at  different  points  of  the  element 
varies  only  by  infinitesimal  amounts ;  cf.  example  («)  below.  By  the 
theorem  of  §  109,  the  infinitesimal  variations  may  be  neglected. 


MULTIPLE  INTEGRALS 


275 


Similarly,   for  a  mass  distributed   along   a  curve   (7,  the 
"linear  density"  is 


g^  Urn  AM^dM^ 
As">.o  As        ds  ' 


and 


M 


=x 


Sds. 


Fig.  97 


Examples :  (a)   Find  the  mass 
of  a  circular  cone  whose  density       q 
varies  as  *  the  distance  from  the 
axis. 

Let  us  take  the  vertex  of  the 
cone   at  the   origin   and  its  axis 
along  OX.     If  we  divide  the  mass 
into  cylindrical  shells  about  the  axis,  each  element  will  be 
"homogeneous"  of  density 

B  =  kr  =  ky. 
We  have 

dV  =  2'7ry{h  —  x)di/^ 

M=fdM=j8dV=2  7rkf"y\h-x)dy 

=  2  7rk  \    y^lh y  \dy  =  -irka^h. 

(5)  The  density  at  any  point  of  a  cube  is  proportional 
to  the  sum  of  the  distances  from  three  adjacent  faces. 
Find  the  mass  of  the  cube. 

Taking  the  three  faces  mentioned  as  coordinate  planes, 
and  choosing  the  element  as  in  §  184,  we  have 

M=  -^  P  P  p(3^  +  ^  H-  z')dz  dy  dx. 


*  To  say  that  a  varies  as  b,  or   a  is  proportional  to  6,  means  that 
a  =  kb,  where  k  is  constant. 


276  CALCULUS 

EXERCISES 

Determine  the  following  masses. 

1.  A  straight  rod  whose  density  is  proportional  to  the  n-th  power 
of  the  distance  from  one  end. 

2.  A  semicircular  wire  whose  density  varies  as  the  distance  from 
the  diameter  joining  the  ends.  Ans.  2  ka^. 

3.  A  circular  plate  whose  density  varies  (a)  as  the  distance  from 
the  center;  (b)  as  the  distance  from  a  fixed  diameter.     Ans.  (b)  f  ka^. 

4.  A  spherical  surface  whose  density  varies  as  the  distance  (a) 
from  a  fixed  diameter  ;   (^>)  from  a  diametral  plane.      Ans.  (a)  kir'^a^. 

5.  A  sphere  whose  density  is  proportional  to  the  distance  from  the 
center.  Ans.  kira^. 

6.  A  rectangle  whose  density  is  proportional  to  the  sum  of  the 
distances  from  two  adjacent  sides. 

7.  A  circular  plate  whose  density  varies  as  the  distance  from  a  point 
on  the  circumference. 

8.  A  square  whose  density  is  proportional  to  the  distance  from  one 
corner.  A  ns.  .765  ka^. 

9.  The  tetrahedron  bounded  by  the  coordinate  planes  and  the 
plane  x  +  y  +  z  =  a,  if  the  density  is  proportional  to  the  sum  of  the 
distances  from  the  coordinate  planes. 

186.  Centroids  and  moments  of  inertia:  the  general  case. 
We  are  now  in  position  to  lay  down  precise  definitions  of 
the  moment  of  first  order,  and  the  moment  of  inertia,  of 
any  mass.  Divide  the  mass  into  elements  AF'as  in  §  184, 
and  multiply  each  element  by  the  density  8  at  a  point 
P  :  (^x,  I/,  z)  of  the  element.  Then  the  moment  of  the  first 
order  with  respect  to  the  ^z-plane  is  defined  as 

lim  yy:VxSAv=  fffxSdv, 

A  K->o  ^^  '^  '^  *^  y^ 

with  a  similar  formula  for  the  moment  with  respect  to  any 
other  plane.  The  centroid  is  defined  as  the  point  (i,  ^,  z) 
whose  coordinates  are  given  by  the  formulas 

Mx  =  fffxhdV,Mi/=fffyhdV,Mz=fffzhdV. 


MULTIPLE  INTEGRALS  277 

Similarly,  the  moment  of  inertia  with  respect  to  any 
axis  is  defined  as 

V 

where  r  is  the  distance  of  a  point  of  the  element  from  the 
axis. 

While  the  above  formulas  are  important  from  the 
theoretical  standpoint  on  account  of  their  generality,  it 
must  not  be  foi'gotten  that  in  the  actual  computation  of 
moments  of  the  first  order  and  moments  of  inertia  multiple 
integrals  are  very  rarely  needed,  at  least  for  homogeneous 
masses. 

It  will  be  remembered  that  the  theorems  of  §  134  have 
been  proved  only  for  a  set  of  particles.  The  reader  will 
now  have  no  difficulty  in  extending  the  proof  to  the 
general  case. 

EXERCISES 

1.  Find  the  centroid  of  the  volume  in  the  first  octant  bounded  by 
the  paraboloid  az  —  x^  +  y'^  and  the  planes  ij  —  x^  x  —  a. 

Ans.    (I a,  ^9^-a,  I'jn). 

2.  Find  the   moment  of  inertia  of  the   volume  in  Ex.  1,  with 
respect  to  the  2:-axis. 

3.  Find  the  centroid  of  the  volume  in  Ex.  3,  p.  262. 

4.  Find  in  two  ways  the  centroid  of  the  volume  in  Ex.  9,  p.  263. 

5.  Find  the  moment  of  inertia,  with  respect  to  the  x-axis,  of  the 
volume  in  Ex.  4,  p.  262.     Check  by  inverting  the  order  of  integration. 

In  Exs.  6-10,  use  polar  coordinates. 

6.  Find  the  moment  of  inertia,  with  respect  to  the  2-axis,  of  the 
volume  in  Ex.  2,  p.  265. 

7.  Find  the  centroid  of  a  hemisphere  (cf.  Ex.  6,  p.  266). 

8.  Find  the  moment  of  inertia  of  a  sphere  about  a  diameter. 

9.  Find  the  centroid  of  a  spherical  wedge  of  half -angle  a.     Check 
by  putting  ct  =  7- 

10.    For  the  wire  of  Ex.  2,  p.  276,  find  (a)  the  centroid ;  also  the 


278  CALCULUS 

moment  of  inertia  with  respect  to  (b)  the  diameter  joining  the  ends, 
(c)  the  radius  perpendicular  to  this  diameter. 

Ans.    (a)  x  =  \Tra;  (b)  ^Ma'^;  (c)  I  Ma^. 

11.  Find  the  moment  of  inertia  of  a  circular  disk  whose  density 
varies  as  the  distance  from  the  center,  (a)  about  the  axis  of  the  disk, 
(b)  about  a  diameter.  Ans.   (a)  ^Ma^. 

12.  Find  the  centroid  of  a  rectangle  whose  density  is  proportional 
to  the  sum  of  the  distances  from  two  adjacent  sides. 

13.  Find  the  moment  of  inertia  with  respect  to  (a)  the  a:^-plane, 
(6)  the  a;-axis,  of  the  volume  bounded  by  the  planes  z  =  x  -\-y,  x+y  =  a, 
and  the  coordinate  planes.  Ans.    (a)  \  Ma^. 

14.  Find  the  moment  of  inertia,  with  respect  to  the  yz-plane,  of  the 
volume  bounded  by  the  planes  x  =  0,  y  =  0,  z  ■=  a,  z  =  x  -^  y,  integrat- 
ing in  the  order  z.,  y,  x\  check  by  integrating  in  the  order  x,  y,  z. 

16.  Find  the  centroid  of  a  straight  rod  whose  density  is  propor- 
tional to  the  n-th  power  of  the  distance  from  one  end. 

.         _     n  -t-  1 , 
Ans.   x= — ■ — I. 
n-h  2 

16.  By  dividing  a  triangle  into  strips  parallel  to  the  base  and  con- 
centrating the  mass  of  each  strip  at  its  center,  show  that  in  finding 
the  centroid  of  the  triangle  we  may  replace  the  triangle  by  a  straight 
line  lying  along  the  median  and  having  a  density  proportional  to  the 
distance  from  the  vertex.  Hence  find  the  centroid  of  any  triangle  by 
the  result  of  Ex.  1.5. 

17.  By  a  method  analogous  to  that  of  Ex.  16,  find  the  centroid  of 
any  cone  or  pyramid. 

18.  Prove  theorems  I  and  IT  of  §  134  for  the  general  case  of  any 
continuous  mass. 


CHAPTER   XXIV 

FLUID    PRESSURE 

187.  Force.  If  a  particle  of  mass  m  moves  with  an  ac- 
celeration y,  the  product  of  the  mass  by  the  acceleration  is 

called  force : 

F=mj\ 
and  the  motion  is  said  to  be  due  to  the  action  of  the  force. 

Since  force  is  a  mere  numerical  multiple  of  acceleration, 
it  follows  that  force  is  a  vector  (§  5(j').  If  several  forces 
act  on  the  same  particle,  their  combined  effects  are  equiva- 
lent to  that  of  a  single  force,  their  resultant.  Usually  the 
resultant  is  most  easily  found  analytically  by  resolving 
each  force  into  components  parallel  to  the  coordinate  axes 
and  summing  in  each  direction  to  get  the  rectangular  com- 
ponents of  the  resultant,  after  which  the  resultant  is  found 
by  compounding  these  rectangular  components  (see  the 
example  below). 

If  there  is  no  force  acting  on  the  particle,  or,  what  is  the 
same  thing,  if  the  resultant  of  all  the  forces  is  0,  the  par- 
ticle is  said  to  be  in  equilibrium.  It  follows  from  §  59 
that  a  particle  in  equilibrium  is  either  at  rest  or  moving 
uniformly  in  a  straight  line. 

If  several  forces  act  at  various  points  of  a  body,  it  is 
not  always  possible  to  compound  them  into  a  single  result- 
ant. In  what  follows,  we  shall  consider  only  cases  in 
which  this  is  possible. 

Example  :     Find  the  resultant  of  a  plane  system  of  forces 

i^i=10  1bs.,   ^2  =20  lbs.,   i^3  =  8  1bs.,   7^^=15  lbs. 

acting  as  in  the  figure,  where  a  =  arctan  |. 

279 


280 


CALCULUS 


Fig.  98 


Hence 


The  components  parallel 
to  OX  are,  of  F^,  10;  of 
F^,  20  cos  a  =16;  of  F^,  0; 
of  F^^  —  15  sin  a  =  —  9. 
"-^  Hence  the  a;-component  Rj. 
of  the  resultant  is 

i2^  =  10  +  16  +  0  -  9  =  17. 

Similarly, 

i^^  =  0  +  12  -  8  -  12  =  -  8. 


B  =  VI72  +  82  =  V353  =  18.8  lbs., 


inclined  to  the  a^-axis  at  an  angle 

arctan(-y\)=-25°12'. 

188.  Force  distributed  over  an  area.  We  have  frequently 
to  consider  a  force  not  acting  at  a  single  point,  but  dis- 
tributed over  an  area.  Examples  are  the  pressure  of  a 
body  of  water  upon  a  dam,  that  of  a  carload  of  sand 
against  the  sides  of  the  car,  the  attraction  of  an  electric 
point-charge  upon  an  electrified  plate,  etc.  If  the  mass 
upon  which  the  force  acts  be  thought  of  as  composed  ulti- 
mately of  particles,  such  a  distributed  force  may  be 
regarded  as  comprising  the  totality  of  forces  acting  on  the 
separate  particles.  We  shall  consider  only  the  case  in 
which  all  these  separate  forces  taken  together  are  equiva- 
lent to  a  single  resultant ;  the  resultant  is  the  total  force 
acting  on  the  body. 

Consider  a  force  acting  in  the  same  direction  at  all 
points  of  a  plane  surface  /S',  and  suppose  for  concreteness 
that  the  force  is  normal  to  the  surface.  If  we  denote  by 
Ai^  the  total  force  acting  on  an  element  of  area  AaS'  chosen 

A  Ti^ 

as  in  §  180,  then  the  ratio  — -  is  called  the  average  pres- 
sure on  ^S.     Now,  if  AaS' approaches  0  in  such  a  way  that  a 


FLUID  PRESSURE 


281 


A    IT 

certain  point  Q  is  always  included,  the  ratio  — -  in  general 

approaches  a  limit,  called  the  pressure  at  the  point  Q: 

T      AF     dF 

V  =    hm   ■ = 

^      A,s'->o  AS     dS 
When  the  pressure  at  every  point  is  given  as  a  func- 
tion of   the  coordinates,  the  total  force  F  can  be  found 
by  integration.     In  the  most  general  case  the  force  ap- 
pears as  a  double  integral,  by  §  180 : 

s 
but  in  most  cases  of  practical  importance  the  element  of 
area  can  be  so  chosen  that  a  single  integration  is  sufficient. 

189.  Fluid  pressure.  An  example  of  force  acting 
normally  to  a  surface  is  furnished  by  the  pressure  of  a 
fluid  against  a  retaining  wall. 

The  pressure,  at  any  point  of  an  incompressible  fluid, 
due  to  the  weight  of  the  fluid  is  equal  to  the  weight  per 
unit  volume  times  the  depth  h  of  the  point  below  the 
surface  of  the  fluid  : 

p  =  wh. 

We  will  assume  the  retaining  area 
to  be  plane  and  vertical.  Let  us 
divide  this  area  into  horizontal 
rectangular  elements  of  area  Z^AA 
as  in  the  figure.  If  we  denote  by 
Pi  the  pressure  at  the  depth  A^, 
the  force  acting  on  the  rectangle 
liAh  is  approximately  * 

pJiAh  =  whiliAh. 

n 

Then  the  sum  ^  whiliAh  is  approximately  the  total  force, 

*  The  actual  force  on  the  rectangle  evidently  differs  from  the  quantity 
PiliAh  by  an  infinitesimal  of  higher  order,  which  may  be  neglected. 


Surface 


Fig.  99 


282 


CALCULUS 


or  total  pressure  *  P,  on  the  whole  area,  and  the  limit  of 
this  sum  is  exactly  P.  Hence,  by  the  fundamental 
theorem  of  §  104, 

where  limits  of  integration  are  to  be  assigned  in  such  a 
way  as  to  extend  the  integration  over  the  whole  area. 

Example :  A  trough,  whose 
cross-section  is  an  equilateral  tri- 
angle of  side  2  ft.,  is  full  of  water. 
Find  the  total  pressure  on  one  end. 
Let  us  take  the  origin  at  the 
lower  vertex  of  the  triangle. 
Then  the  equation  of  the  line 
OA  is 

^  =  VS  X. 
The  total  pressure  on  the  triangle  is 


dx 


P  =  2wf  "( V3  -  ^)x  d^  =  2  wJ\VS  -  V§  x)x^S 

=  6  w  \    (1  —  x^x  dx=  Q  tv\ 

=  w  =  62  lbs.,  nearly. 

EXERCISES 

1.  A  particle  is  acted  on  by  two  forces  F^,  F^  lying  in  the  same 
vertical  plane  and  inclined  to  the  horizon  at  angles  ctj,  a^.  Find  their 
resultant  in  magnitude  and  direction,  if  F^  =  527  lbs.,  F^  =  272  lbs., 
«!  =  127°  52',  a2  =  32°  13'. 

Ans.    569  lbs.,  inclined  to  the  horizon  at  99°  26'. 

2.  Six  forces,  of  1,  2,  3,  4,  5,  6  lbs.  respectively,  act  at  the  same 
point,  making  angles  of  60°  with  each  other.     Find  their  resultant. 

Ans.    6  lbs.,  acting  along  the  line  of  the  5-lb.  force. 

*  Care  must  be  taken  not  to  confuse  the  total  pressure^  P,  with  the 
pressure  at  a  point.,  p.  The  former  is  a  force,  the  latter,  a  force  per  unit 
area. 


FLUID  PRESSURE  283 

3.  Work  the  example  of  §  189  with  the  origin  at  B. 

4.  Find  the  total  pressure  on  one  side  of  a  plank  2  x  8  ft.  sub- 
merged vertically  with  its  upper  end  (a)  in  the  surface,  (h)  4  ft. 
below  the  surface. 

5.  A  horizontal  cylindrical  boiler  4  ft.  in  diameter  is  half  full  of 
water.     Find  the  total  pressure  on  one  end.  Ans.   330  lbs. 

6.  Work  Ex.  5  if  the  boiler  is  full  of  water. 

7.  What  force  must  be  withstood  by  a  vertical  dam  100  ft.  long 
and  20  ft.  deep? 

8.  Work  Ex.  7  if  the  dam  is  a  trapezoid  100  ft.  long  at  the  top 
and  80  ft.  long  at  the  bottom,  taking  the  origin  at  an  upper  corner. 
Check  by  solving  again  with  the  origin  in  a  different  position. 

9.  Find  the  total  pressure  on  one  side  of  a  right  triangle  of  sides 
AB  =  3  ft.,  AC  =  4  ft.,  submerged  with  A C  vertical  and  (a)  A  in  the 
surface,  (h)  A  2  ft.  deep,  (c)  C  2  ft.  deep.  In  each  case  check  as  in 
Ex.  8. 

10.  Find  the  total  pressure  on  one  face  of  a  square  2  ft.  on  a  side, 
submerged  with  one  diagonal  vertical  and  one  corner  in  the  surface. 

11.  Find  the  force  on  one  end  of  a  parabolic  trough  full  of  water, 
if  the  depth  is  2  ft.  and  the  width  across  the  top  2  ft.  Ans.  ff  w. 

12.  A  trough  4  ft.  deep  and  6  ft.  wide  has  semi-elliptical  ends.  If 
the  trough  is  full  of  water,  find  the  pressure  on  one  end. 

13.  Find  the  force  that  must  be  withstood  by  a  bulkhead  closing  a 
watermain  4  ft.  in  diameter,  if  the  surface  of  the  water  in  the  reservoir 
is  40  ft.  above  the  center  of  the  bulkhead.  Ans.   16  tons, 

14.  Show  that  the  problem  of  §  189  is  analytically  equivalent  to  the 
following :  To  find  the  mass  of  a  thin  plate,  if  the  density  is  propor- 
tional to  the  distance  from  a  line  in  the  plane  of  the  plate. 

190.  Resultant  of  parallel  forces.  Suppose  we  have 
given  a  set  of  parallel  forces  /j,  f^,  •■•-,fn-,  whose  resultant 
(algebraic  sum)  F  is  not  0.  The  problem  of  finding  the 
line  of  action  of  the  resultant  is  analogous  to  that  of  find- 
ing the  centroid  of  a  set  of  mass  particles. 

Let  us  take  the  a;«/-plane  perpendicular  to  the  given 
forces,  and  let  (rr.-, «/,)  be  the  point  where  the  line  of 
action  of /<  pierces  this  plane.  The  moment  of  the  result- 
ant about  each  coordinate  axis  must  equal  the  sum  of  the 


284  CALCULUS 

moments  of  the  forces  about  the  same  axis.      Hence  the 
line  of  action  of  ^pierces  the  a;?/-plane  at  the  point  whose' 
coordinates  ^,  'y  are  given  by  the  formulas 

n  n 

Fx  =  ^fiXi,    Fy  =  ^fiVi. 

191.  Center  of  pressure.  More  generally,  consider  again 
the  case  of  a  force  acting  normally  at  all  points  of  a  plane 
area.  Take  the  given  plane  as  rr^z-plane,  and  divide  the 
surface  into  elements  A/Sas  in  §  180  ;  then  the  force  on  AaS* 
is  approximately  p^S^  where  p  is  the  pressure  at  a  point 
(a:,  y)  of  A/S^,  and  the  moment  of  this  force  about  the 
«/-axis  is  xpAS.  The  sum  of  these  moments  is  approxi- 
mately the  moment  of  the  whole  force,  and  the  limit  of  the 
sum  is  exactly  that  moment.  Similarly,  we  can  find  the 
moment  about  the  2:-axis.  Hence  the  resultant  acts  at 
the  point  whose  coordinates  ^,  ^  are  given  by  the  formulas 

Fx=ffxpdS,   Fy^ffypdS, 

*  s  s 

where  F  is  the  total  force.     The  point  (x^  ^)   is  called 
the  center  of  pressure. 

As  usual,  it  happens  in  many  problems  that  the  double 
integrals  reduce  to  simple  integrals,  if  the  element  be 
properly  chosen.  In  particular,  in  the  problem  of  fluid 
pressure  it  is  easily  seen  that  the  depth  of  the  center  of 
pressure  below  the  surface  is  given  by  the  formula 

Ph  =  tv^hH  dh, 
where  P  is  the  total  pressure. 

EXERCISES 

1.  A  straight  beam  AB  50  ft.  long  bears  loads  as  follows:  100 
lbs.  at  A,  100  lbs.  at  C,  200  lbs.  at  A  50  lbs.  at  B;  AC  =  10  ft., 
AD  =  20  ft.     Find  the  point  of  application  of  the  resultant. 

•  2.    Work  Ex.  1  if  the  segment  AD  bears  a  uniformly  distributed 
load  of  5  lbs.  per  foot. 


FLUID  PRESSURE  285 

3.  Work  Ex.  2  if  the  segment  DB  bears  a  distributed  load  which 
increases  uniformly  from  5  lbs.  per  foot  at  D  to  15  lbs.  per  foot  at  B. 

4.  A  platform  ABCD  20  ft.  square  bears  a  single  concentrated 
load.  The  reactions  are,  at  A,  50  lbs.;  at  B^  80  lbs.;  at  C,  100  lbs. ; 
at  D,  70  lbs.     Where  is  the  load  ? 

5.  Find  the  most  advantageous  length  for  a  lever  to  lift  a  weight 
of  100  lbs.,  if  the  distance  from  the  weight  to  the  fulcrum  is  4  ft. 
and  the  lever  weighs  4  lbs.  per  foot. 

Find  the  depth  of  the  center  of  pressure  in  the  following  cases. 

6.  A  rectangle  submerged  vertically  (a)  with  one  edge  in  the 
surface,  (h)  with  its  upper  edge  at  a  depth  c.  Ans.  («)  f  a. 

7.  An  isosceles  triangle  submerged  with  the  line  of  symmetry 
vertical  and  (a)  the  vertex,  (6)  the  base,  in  the  surface. 

Ans.  (a)  I  A;  (6)  \  h. 

8.  Any  triangle  submerged  with  one  side  in  the  surface. 

9.  One  end  of  the  parabolic  trough  of  Ex.  11,  p.  283. 

10.  A  semicircle  submerged  with  its  bounding  diameter  in  the 
surface. 

11.  In  each  case  of  Ex.  9,  p.  283,  if  the  pressure  is  removed  from 
one  side  of  the  triangle,  at  what  point  must  a  brace  be  applied  in 
order  to  hold  the  triangle  in  position  ? 

.4ns.  (a)  With  AB,  AC  as  axes,  (|,  2). 

12.  Show  that  the  problem  of  §  191  is  analytically  equivalent  to 
that  of  finding  the  centroid  of  a  plane  mass  of  variable  density  p. 


CHAPTER   XXV 
DIFFERENTIAL  EQUATIONS   OF   THE    FIRST   ORDER 

I.    General  Introduction 

192.  Differential  equations.  A  differential  equation  is 
an  equation  that  involves  derivatives  or  differentials. 
Various  examples  have  arisen  in  our  previous  work,  of 
which  the  following  may  be  mentioned : 

(§15) 

(§16) 
(§50) 

(Ex.  43,  p.  64) 
(§51) 

(Ex.  18,  p.  50) 

(Ex.  1,  p.  81) 

(Ex.  2,  p.  78) 

(Ex.  1,  p.  2.39) 

(10)  ?1  +  f^  =  0-  (Ex.  22,  p.  240) 

Equations  containing  partial  derivatives,  such  as  ex- 
amples (9)  and  (10),  are  called  partial  differential  equa- 
tions. Such  equations  are  of  great  importance,  but  a 
study  of  them  is  beyond  the  limits  of  this  book. 

286 


(1) 

ds          1 

dt          t^ 

(2) 

y"=4. 

(3) 

dy  =  2  cos  2  6  d6. 

W 

y{n)   ^  ^n^ax^ 

(5) 

xdx  +  y  dy  =  ^, 

(6) 

(7) 

(8) 

(i+y¥  =  „. 

y 

(9) 

f  =  2^  +  2/-.3. 
ox 

DIFFERENTIAL  EQUATIONS  OF  FIRST  ORDER     287 

193.  Order  of  a  differential  equation.  The  order  of  a 
differential  equation  is  the  order  of  the  highest  derivative 
that  occurs  in  it.  Tlius,  in  §  192,  examples  (1),  (3),  (5) 
are  of  the  first  order,  (2),  (6),  (7),  (8)  are  of  the  second 
order,  (4)  is  of  the  n-th  order. 

In  the  applications,  equations  of  the  first  and  second 
orders  are  of  predominant  importance,  and  we  shall  be 
chiefly  concerned  with  these  two  types. 

194.  Solutions  of  a  differential  equation.  A  solution  of 
a  differential  equation  is  any  relation  between  x  and  y 
by  virtue  of  which  the  differential  equation  is  satisfied. 
Thus  equation  (1)  of  §  192  is  true  if 

1^ 

where  c  is  arbitrary ;  hence  this  relation  is  a  solution  of 
the  equation.     A  solution  of  (2)  is  easily  seen  to  be 

It  appears  from  these  examples  that  a  solution  of  a 
differential  equation  may  involve  one  or  more  arbitrary 
constants ;  we  shall  find  this  to  be  true  in  general.  It 
follows  that  each  equation  has  an  infinity  of  solutions, 
obtained  by  assigning  different  values  to  the  arbitrary 
constants. 

By  analogy  with  the  integral  calculus,  a  solution  of  a 
differential  equation  is  often  called  an  integral  of  the 
equation,  and  the  arbitrary  constants  are  called  constants 
of  integration. 

II.    Equations  of  the  First  Order 

195.  The  general  solution.  Suppose  there  is  given  a 
relation  (free  of  derivatives)  between  x,  y  and  an  arbi- 
trary constant : 

(1)  Fix,  y,  C-)  =  0. 


288  CALCULUS 

Geometrically  this  equation  represents  a  family  of  curves^ 
whose  individual  members  are  obtained  by  assigning  par- 
ticular values  to  c. 

If  we  differentiate  (1)  with  respect  to  x^  the  arbitrary 
constant  c  may  be  eliminated  from  the  equation  thus 
formed  and  the  original  equation.  The  result  of  this 
elimination  is  evidently  an  equation  involving  x^  y^  and 
y^ ;  i,e.  it  is  a  differential  equation  of  the  first  order  : 

(2)    ■  *Cr^,y,j/')=0. 

As  this  equation  does  not  contain  c,  it  represents  a  prop- 
erty common  to  all  the  curves  of  the  above-mentioned 
family. 

Since  equation  (2)  is  true  by  virtue  of  equation  (1),  it 
follows  that  (1)  is  a  solution  of  (2). 

If  a  solution  of  a  differential  equation  of  the  first  order 
contains  an  arbitrary  constant,  it  is  called  the  general 
solution:  hence  (1)  is  the  general  solution  of  (2).  It 
can  be  shown  that,  in  general,  corresponding  to  every 
differential  equation  of  the  form  (2)  there  exists  a  gen- 
eral solution  (1)  ;  methods  of  finding  this  solution  in 
various  cases  will  be  considered  presently. 

It  may  be  worth  while  to  point  out  that,  if  the  differ- 
ential equation  has  the  simple  form 

ax 
the   integral    calculus   gives   us   the  general   solution   at 


once 


y=jf(x)dx  +  c. 


It  should  also  be  noted  that  while  in  the  integral  cal- 
culus the  constant  of  integration  always  appears  as  an 
additive  constant,  this  is  not  true  in  general  in  the  solu- 
tion of  a  differential  equation ;  the  constant  often  enters 
in  other  ways. 


DIFFERENTIAL  EQUATIONS  OF  FIRST  ORDER     289 

Examples :  (a)  Find  the  differential  equation  whose 
general  solution  is 

Differentiating,  we  find 

di/  =  2  ce^  dx ; 

eliminating  c  by  division,  we  get 

^  =  2dx. 

y 

This  example  illustrates  the  fact  that  the  arbitrary  con- 
stant is  not  always  additive. 

(5)  Find    by    inspection    the    general    solution    of   the 

equation 

X  dy  •\-  y  dx=(). 

The  answer  is  seen  at  once  to  be 

xy  =  e. 

196.  Particular  solutions.  A  solution  obtained  from 
the  general  solution  by  assigning  a  particular  value  to 
the  arbitrary  constant  is  called  a  particular  solution  of  the 
differential  equation.  Thus  in  example  (^),  §  195,  the 
equations   xy  =0,  xy  =  5,  etc.,    are    particular   solutions. 

In  applied  problems  involving  differential  equations 
we  are  often  concerned  with  a  particular  solution. 
Nevertheless  the  determination  of  the  general  solution 
is  usually  a  necessary  preliminary  step,  after  which  the 
required  particular  solution  is  found  by  determining 
the  arbitrary  constant  from  given  initial  conditions.  The 
process  is  illustrated  by  the  examples  of  §  77,  which 
should  be  reviewed  at  this  point. 

Differential  equations  involving  y'  to  a  degree  higher 
than  the  first  may  in  some  cases  have  a  so-called  singular 
solution.,  which  cannot  be  obtained  from  the  general  solu- 
tion by  assigning  a  particular  value  to  the  arbitrary  con- 
stant. As  such  solutions  are  of  little  importance  in  most 
of  the  elementary  applications,  we  shall  omit  a  discussion 
of  them. 


290  CALCULUS 


EXERCISES 

In  the  following  cases,  find  the  differential  equation  whose  general 
solution  is  the  given  equation. 

1.    y  =  x^  -{-  c.  2.   y  —  ex. 

Z.   y  =  ce^.  A.   y  =  ex  -\-  c^. 

6.  log  r  =  kO.  6.    xy  -\-  cy  =  \. 

7.  s  =  sin  t  -f  c  cos  t.  8.    c^  +  2  ey  =  x^. 

Find  by  inspection  the  general  solution  of  each  of  the  following 
differential  equations. 

9.    dy  —  sin  x  dx  =  0.  10.    x  dx  -\-  y  dy  =  d. 

11.  ^=xdx.  12.    ^i^  =  ^'. 

y  y       X 

13.  xdy  +  y  dx  -\-  2  dy  =  0. 

14.  Find  the  equation  of  a  curve  M'hose  slope  at  any  point  is 
equal*  to  the  abscissa  of  the  point.  How  many  such  curves  are 
there?     Draw  several  of  them. 

15.  In  Ex.  14,  find  the  curve  that  passes  through  (4,  —  3). 

16.  Solve  Ex.  14,  reading  "  ordinate  "  instead  of  "  abscissa." 

17.  A  point,  starting  with  a  velocity  of  10  ft.  per  second,  moves 
under  a  constant  acceleration  of  8  ft.  per  second  per  second.  Find 
(a)  the  velocity,  (/>)  the  distance  from  the  starting  point,  after  t 
seconds  of  motion. 

18.  A  point  moves  under  an  acceleration 

dv  4         o  * 

—  =  —  4  cos  2  t. 

dt 
If  17  =  0  and  X  =  1  when  ^  =  0,  find  v  and  x  in  terms  of  t. 

197.  Geometrical  interpretation.  In  analytic  geometry 
we  find  that  the  locus  of  a  point  whose  coordinates  x^  y  are 
connected  by  an  equation 

is  a  certain  curve,  the  graph  of  the  equation.  In  general, 
any  value  whatever  may  be  assigned  to  x^  and  the  corre- 
sponding value  of  y  determined. 

*  That  is,  the  number  representing  the  slope  is  the  same  as  that  rep- 
resenting the  abscissa.  It  is  only  in  this  sense  that  a  ratio,  such  as 
the  slope  of  a  curve,  can  be  equal  to  a  length,  such  as  the  abscissa  of  a 
point. 


DIFFERENTIAL  EQUATIONS  OF  FIRST  ORDER     291 

If  now  we  have  given  a  differential  equation  of  the 
first  order,  and  of  the  first  degree  in  y\  i.e.  a  relation 
between  a;,  y^  and  y'  of  the  form 

(1)  y'  =  Fix,  y-), 

it  is  clear  that,  in  general,  any  values  whatever  may  be 
assigned  to  x  and  y  provided  we  associate  with  them  the 
value  of  «/'  given  by  the  equation.  Thus,  equation  (1)  is 
satisfied  by  the  coordinates  of  any  point  (x^  y)  provided 
the  point  is  moving  in  the  proper  direction.  Starting  with 
any  assumed  initial  position,  and  moving  always  in  the 
direction  required  by  the  given  equation,  the  point  de- 
scribes a  curve ;  the  values  of  a;,  y^  y'  at  any  point  of  the 
curve  satisfy  the  differential  equation.  Further,  since 
the  initial  position  is  entirely  arbitrary,  it  is  clear  that 
the  point  may  be  made  to  describe  any  one  of  a  family  of 
curves,  the  so-called  integral  curves.  The  equation  of  this 
family  is,  of  course,  the  general  solution  of  the  differ- 
ential equation  ;  it  contains,  as  it  should,  an  arbitrary 
parameter,  viz.,  the  constant  of  integration.  The  graph 
of  any  particular  solution  is  merely  one  of  the  family  of 
integral  curves. 

Example  ;  Interpret  geometrically  the  differential  equa- 
tion 

X  dx  -i-y  dy  =  0. 

Writing  the  equation  in  the  form 

dy  _  _x 

dx  y 

we  see  that  the  point  (a:,  y')  must  always  be  moving  in  a 
direction  perpendicular  to  the  line  joining  it  to  the  origin. 
Its  path  is  therefore  any  one  of  the  family  of  circles  with 
center  at  the  origin.  This  may  be  verified  by  observing 
that  the  general  solution  of  the  differential  equation  is 

2^  -{-  y^  =  c. 


292  CALCULUS 

EXERCISES 

In  each  of  the  following  cases  find  the  equation  of  the  family  of 
integral  curves  and  draw  several  curves  of  the  family. 

1.   ^f  =  0.  2.   y<  =  5. 

^  ax     X 

5.    ^  =  -y.  6.  y'  =  y. 

dx  X 

7.  Find  the  differential  equation  of  the  family  of  circles  through 
the  origin  with  centers  on  the  x-axis.  A  ns.     2  xyy'  =  y"^  —  x^. 

8.  Find  the  differential  equation  of  the  family  of  parabolas  with 
foci  at  the  origin  and  axes  coinciding  with  the  x-axis. 

9.  Interpret  geometrically  the  equations  in  Exs.  9,  12,  and  14, 
p.  290. 

198.  Separation  of  variables.  In  the  remainder  of  this 
chapter  we  show  how  to  find  the  general  solution  of  a 
differential  equation  of  the  first  order  in  some  of  the 
simpler  cases. 

Every  differential  equation  of  the  first  order,  and  of  the 
first  degree  in  «/',  can  evidently  be  written  in  the  form 

Mdx  +  iVc?y  =  0, 

where  in  general  il!f  and  iVare  functions  of  both  x  and  t/. 
It  is  often  possible  to  transform  the  equation  so  that  Jf  is 
a  function  of  x  alone  and  iVis  a  function  of  ^  alone  ;  this 
transformation  is  called  separation  of  variables.  When 
the  variables  have  been  separated,  the  differential  equa- 
tion may  be  solved  by  a  simple  integration,  as  in  the  fol- 
lowing 

Example :  Solve  the  equation 

XT/  dx  -j-  (a;^  +  1)  c?^  =  0. 

After   division   by  y(x^  +  1)    the    equation   takes   the 

form 

xdx    ,dy_r. 


DIFFERENTIAL  EQUATIONS  OF  FIRST  ORDER     293 
Integrating,  we  get 

ll0g(2;2+  1)  +  log?/  =  (7, 

or 


log  y^x^  -f-  1  =  c, 
«/Va;2  -f  1  =  gc^ 
y\x^  +  1)  =  c\ 
where  ^'  =  e^c^ 

EXERCISES 

Solve  the  following  differential  equations. 

1.  (1  +  x)y  dx  +  (1  —  y)x  dy  =  0.  Ans.  log  {xy)  +  x  —  y  =  c. 

2.  y'  =  axy'^.  Ans.    ax~y  +  cy  +  2  =  0. 

3.  &mx  cosy  dx  ■=  co^  X  s,\n  y  dy.  Ans.   cosy— c  cos  x. 

4. -^  +  /zza2.  Ans.   ?L±^  =  ce^a., 

dx  y  —  Cb 


1  +  y      1  -  x 
6.    (1  +  x)y^dx  —  x^  dy  =  0. 


7.    Vl  -  2/2c^x  +  Vl  -x2f/2/ =  0.     J^ns.   xVl  -  y^ -^  yVl 


x"-  =  c. 


8.  '^=-kv^. 
dt 

9.  ^  =  _  cos  2  t. 
dt 

10.  Show  that  the  function 

y  =  ce* 

is  the  only  function  that  is  unchanged  by  differentiation. 

11.  Find  a  function  whose  first  derivative  is  equal  to  the  square  of 
the  original  function.*     Interpret  geometrically. 

12.  Determine  the  family  of  curves  whose  slope  at  any  point  is 
equal  to  the  product  of  the  coordinates  of  the  point.  Find  the  curve 
of  this  family  that  passes  through  the  point  (0,  1),  and  trace  it. 

13.  A  particle  falls  under  gravity,  the  resistance  of  the  air  being 
neglected.     If  the  initial  velocity  is  y^,  find  y  and  a;  in  terms  of  t. 

*  Cf .  footnote,  p.  290. 


294  CALCULUS 

14.  Determine  the  family  of  curves  represented  by  the  equation 

dx 

15.  In  Ex.  14,  find  the  curve  (a)  that  passes  through  (0,  0) ;   (&) 
that  crosses  the  line  a;  =  1  at  an  angle  of  45°.     Trace  these  curves. 

199.    Coefficients  homogeneous  of   the   same   degree.     A 

polynomial  in  x  and  y  is  said  to  be  homogeneous  if  all  the 
terms  are  of  the  same  degree  in  x  and  y.  More  generally, 
an}^  function  of  x  and  y  is  said  to  be  homogeneous  of  the  n-th 
degree  if,  when  x  and  y  are  replaced  by  kx  and  ky  respec- 
tively, the  result  is  the  original  function  multiplied  by  k"^. 
Thus  the  function 

X  +  Va;^  —  y'^  -\-  y  log  ^ 

X 

is  homogeneous  of  the  first  degree. 
If,  in  the  equation 

Mdx  +  Ndy  =  0, 

the  coefficients  iHfand  iVare  homogeneous  functions  of  the 
same  degree^  it  is  easily  seen  that  the  equation  when  solved 
for  y^  takes  the  form 

i.e.  y'  is  a  function  of  ^  alone.     This  suggests  the  substitu- 

X 

tion  of  a  new  variable  v  for  the  ratio  ^;  i.e.  the  substitu- 

..  X 

tion 

y  =  vx.,  dy  =  V  dx  -\-  X  dv. 

This  substitution  always  produces  a  differential  equation 
in  V  and  x  in  which  the  variables  are  separable. 
Example  :  Solve  the  equation 

(rr  -\-  y)dx  —  x  dy  =  ^. 
Substituting 

y=  vx^  dy  =  V  dx  +  X  dv., 
we  find 

(ic  -H  vx)dx  —  x(y  dx  -\-  x  dv)  =  0, 
or 

dx  —  xdv  =  ^. 


DIFFERENTIAL  EQUATIONS  OF  FIRST  ORDER     295 


The  variables 

can  ] 

now  be  separated  : 

dx 

-  dv=0, 

X 

log 

X  —  V  -\-  c  =  ^^ 

or,  since 

V  =^^ 

X 

y  =^  X  log  X  -\-  ex. 

EXERCISES 
Solve  the  following  differential  equations. 

1.  {x  +  y)y'  -}-  X  —  y=  0.        Ans.    arctan  ^  +  -  log  (x^  +  y^)  =  c. 

X      2 

2.  (^x'^  +  y^)dx  —  2  xy  dy  =  0.  Ans.   x-  —  y^  =  ex. 

du  *- 

3.  (xy  —  x^)—  =  y'^.  Ans.    y  =  ce'. 

dx 

4.  x^  dy  +  y^  dx  =  0. 
6.    x^  dx  +  y^  dy  =  0. 


6.    u  dv  —  V  du  —  Vu'^  4-  v'^  du  =  0.  Ans.    u^  =  c^  +  2  cv. 


7.  x  dx  +  Va;"^  +  1  ^^  =  0. 

8.  2  uv  du  +  (^2  -  3  u^)dv  =  0.  Ans.    v^  =  c(u^  -  v^). 

9.  V  —  =  —  X  —  V. 

dx 

10.  Show  that,  if  AI  and  N  are  homogeneous  of  the  same  degree, 
the  equation 

M  dx  +  N  dy  =0 

can  always  be  put  in  the  form 

11.  Give  a  general  proof  of  the  fact  that,  in  the  problem  of  §  199, 
the  substitution  y  =  vx  always  leads  to  an  equation  in  which  the 
variables  are  separable. 

200.  Exact  differentials.  The  differential  of  a  function 
u  of  two  variables  x  and  y  is  given  by  formula  (1)  of 
§164: 

(1)  du  =  —  dx  -\ dy. 

dx  dy 


296  CALCULUS 

The  quantity 

(2)  Mdx  +  Ndy 

is  called  an  exact  differential  if  it  is  precisely  the  dif- 
ferential of  some  function  u.  Thus,  the  quantity 
X  dy  •\-  y  dx  is  an  exact  differential,  viz.  d(xy^ ;  on  the 
other  hand,  the  quantity  x  dy  —  y  dx  is  not  an  exact 
differential. 

If  the  quantity  (2)  is  an  exact  differential,  it  appears 
by  comparison  with  (1)  that  there  must  exist  a  function 
u  such  that 

(3)  ^  =  M, 

dx 

(4)  ^^  =  K 

dy 

Differentiating  (3)  with  respect  to  y  and  (4)  with  respect 
to  x^  we  find 

J2^  ^  QM       BH   _  dN 

dy  dx       dy       dxdy       dx 

c)lJI  filJf 

Equating  values  of  - — —  and  -,  by  §  163,  we  get  the 

dy  dx  dx  dy 

relation 

dM^dJsr 

dy        dx 

as  a  necessary  condition  that  (2)  be  an  exact  differential. 
It  can  be  shown  that  this  condition  is  not  only  necessary 
but  sufficient:  i.e.  the  quantity  M  dx -\- N  dy  is  an  exact 
differential  if  and  only  if 

dM  ^dN 
dy  ~  dx' 

201.    Exact  differential  equations.     The  equation 

(1)  Mdx-\-Ndy=Q 

is  called  an  exact  differential  equation  if  its  left  member  is 
an  exact  differential. 


DIFFERENTIAL  EQUATIONS  OF  FIRST  ORDER     297 

Since  equation  (1),  when  exact,  has  the  form 

du  =  0, 
its  general  solution  is  evidently 

u  =  c. 

While  a   general   method  can    be   given   for  finding  the 
function  u^  we  shall   consider    only  cases  in    which   this 
function  is  readily  found  by  inspection. 
202.    Integrating  factors.     If  the  equation 

(1)  Mdx-\-Ndi/  =  0 

is  not  exact,  its  solution  can  still  be  put  in  the  form 

(2)  u  =  c 

by  merely  solving  for  the  arbitrary  constant.  By  differ- 
entiating (2)  we  obtain  an  equation  of  the  first  order  that 
is  satisfied  whenever  (1)  is  satisfied:  this  equation  must 
therefore  have  the  form 

v(Mdx-\-]Srdi/}=0, 

where  v  is  in  general  a  function  of  both  x  and  y.  Thus 
for  every  differential  equation*  (1)  there  exists  a  function 
V,  called  an  integratiiig  factor^  whose  introduction  renders 
the  equation  exact. 

It  can  be  shown  that  every  differential  equation  has 
not  merely  one,  but  infinitely  many,  integrating  factors  ; 
nevertheless  it  is  frequently  impossible  to  find  one  of 
them.  In  various  cases,  some  of  which  will  be  considered 
presently,  an  integrating  factor  can  be  found  by  direct 
processes  ;  in  other  cases  it  is  best  found  by  inspection. 

It  should  be  noticed  that  in  separating  variables,  as  in 
§  198,  we  are  really  introducing  an  integrating  factor. 
Thus,  in  the  example  of  that  article,  the  integrating 
factor  is 

1 
K^  +  1)' 

*  Assuming  the  existence  of  the  general  solution.     Cf.  §  195. 


298  '  CALCULUS 

Example  :  Solve  the  differential  equation 
X  dy  —  y  dx  =^  0. 

If  we  note  thalt  the  differential  of  ^  is  ^  ^V  -  V  ^^  ^  it 

X  x^ 

appears  that  —  is  an  integrating   factor  in  the  present 
x^ 

instance  : 

X  dy  —  y  dx  _  r. 


x^ 

x 

y  =  ex. 

Other  integrating   factors  are  —  (which  merely  sepa- 

rates  the  variables),  — ,    — -• 

y^      x^  ±  y^ 

EXERCISES 

1.  Solve   the   above  example   by  using  each  of   the  integrating 
factors  there  mentioned,  and  compare  the  results. 

2.  Solve  Ex.  1,  p.  295,  by  means  of  an  integrating  factor. 

Solve  the  following  equations. 

3.  xdy  —(x  -\-  y)  dx  =  0. 

4.  (2  a:  +  2  ?/)  dx  +  (2x  +  f)dy  =  0. 

5.  (x  —  y^)  dx  +  2  xy  dy  =  0. 

6.  xdy  —  y  dx  =  (x^  +  y'^)  dx. 

7.  (a:  +  y  +  1)  dx  -\- (x  -  y)  dy  =  0. 

8.  xdx  -^  y  dy  +  x  dy  —  y  dx  =  0. 


9. 

xy'  =  y  -\-  Vx^  —  y'^. 

10. 

u(u  +  2  v)  du  +  (m2  -  v^)  dv 

=  0. 

11. 

^dv__  1 
^ds          s^' 

12. 

(sin  y  +  2  x)  dx  -\-  X  cos  y  dy 

=  0. 

203.  The  linear  equation.  A  differential  equation  of 
the  tirst  order  is  said  to  be  linear  if  it  is  of  the  first  degree 
in   y   and   y' ,     Every  such    equation   may  evidently    be 


DIFFERENTIAL  EQUATIONS  OF  FIRST  ORDER     299 
written  in  the  form 

(1)  y'  +Py=-  Q, 

where  P  and  Q  are  functions  of  x  alone.  We  shall  find 
that  the  linear  equation  is  of  especial  importance  in  the 
applications. 

Before  undertaking  to  solve  equation  (1),  let  us  con- 
sider the  special  case 

(2)  /  +  P2/=0. 

Here  the  variables  are  separable,  and  the  solution  may 
be  obtained  at  once  : 

^-{-Pdx  =  0, 

y 

whence 

log  y  +  \  P  dx  =  c^ 

(3)  ye^^'^-"  =  c'. 
Now,  differentiating  (3),  we  get 

elPdx(^^y  _|_  py  ^^^  ^  Q^ 

which  shows  that  e^^'^'^  is  an  integrating  factor  for  equa- 
tion (2).     But  since  Q  is  a  function  of  x  alone,  it  follows 
that  e^^^"^  is  likewise  an  integrating  factor  for  equation  (1). 
Examples :  (a)   Solve  the  equation 

dy  -\-^y  dx=  x  dx. 
Here 

P=2,    CPdx  =  2x,  e^^'^'^  =  e^. 

Introducing  the  integrating  factor  e^^^  and  integrating, 
we  find 

ye^  =  I  xe^""  dx  ==  I-  xe^""  —  je^""  -\-  (?, 

whence 

y  =  \x  —  \-\-  ce-2^. 

(6)  Solve  the  equation 

xy'  —  x^  —  y  =  0. 

Writing  this  in  the  form 

(4)  dy  —  ^dx  =  x^dx, 

X 


300  CALCULUS 

we  have 

P  = ,    ]  P  dx=  —  logx^ 

X     ^ 

whence 

X 

by  formula  (5)  of  §  44.  Hence,  dividing  equation  (4) 
by  X  and  integrating,  we  get 

^  =  i  xdx  =  — \-  c, 
X     *^  2 

2^  =  a^  -\-  c'x. 

204.    Equations  linear  in/(i/).     The  equation 

(1)  /w+^/w=e, 

where  P  and  Q  are  functions  of  x  alone,  is  evidently 
linear  in  /(«/),  and  may  be  solved  by  the  method  of  the 
preceding  article. 

An  equation  not  given  directly  in  the  form  (1)  may 
sometimes  be  reduced  to  that  form  by  a  simple  trans- 
formation. In  particular,  this  is  always  possible  with  the 
equation. 

The  process  is  as  shown  in  the  following 
Example:  Solve  the  equation 

X     y^ 
Let  us  write  the  equation  in  the  form 

y^  dy  -\-'^dx=  dx. 

X 

If  we  multiply  through  by  3,  so  that  the  first  term  be- 
comes c?(^^),  this  equation  is  seen  to  be  linear  in  y^ : 

3  v^ 
Sy^dy  -i — ^  dx=S  dx. 

X 

Here 

P  =z-      e^^  ^^  =  g31og  X  _  ^ 
X 


DIFFERENTIAL  EQUATIONS  OF  FIRST  ORDER  301 
whence  the  solution  of  the  equation  is 

^3^3  _  3  i  2'^  dx  =  I  x'^  -{-  c, 

^3  _  I  ^  _j_  cx~^. 

EXERCISES 

Solve  the  following  equations. 

1.  -^-\-  y  =  X.  Ans.   y  =  x  —  1  +  cc*. 
dx 

2.  {x  +  \)  dy  -  2y  dx  =(x  +  \)^dx. 

Ans.   2y  =(x  +  ly  -\-  c(x  +  l)^. 

3.  y'  -  xy  =  X. 

4.  X—  +  (1  -\-  x)y  =  e^.  Atis.   2  x?/e^  =  e'^  +  c. 

dx 

5.  (x  -  2y  +  5)  dx  -j- (2  x  +  4:)dy  =  0. 

6.  y'  sin  y  +  sin  x  cos  y  =  sin  x. 

7.  dy  +  1/(1  —  xy^)  dx  =  0.  Ans.    —  =  a:  H h  ce^. 

8.  ^y^y'  —  2//^  =  X  -f  1.  ^ns.    ?/^  =  ce^  —  2  ^  —  !• 

9.  -—  =  g  —  kv.     Solve  in  two  ways. 

10.  — ^  cos  X  -\-  y  sin  x  =  1. 
c/a: 

11.  —+  y  cos  a:  =  sin  2  x. 

12.  (1  +  a;2)  -^  -I-  ?/  =  arctan  x. 

dx 

13.  —  z=  —  V  +  cos  ^ 

14.  (xy"^  +  ?/)f/a:  —  x  dy  =  0. 

15.  dydx  +  (x  +  a:?/2)  f//y  =  0. 

16.  ydy  +  (xy^  —  x)  dx  =  0.     Solve  in  two  ways. 

17.  X  dy  +  {xey  -\)dx  —  ^. 

205.  Geometric  applications.  Many  of  the  properties 
of  a  curve  depend  not  only  on  the  coordinates  x^  y^  but 
on  the  slope  y^  as  well.     When  a  curve  is  defined  by  such 


302  CALCULUS 

properties,  the  analytic  expression  of  the  given  data  leads 
to  a  relation  between  x^  y,  and  y'  —  in  other  words,  to  a 
differential  equation  of  the  first  order.  The  general 
solution  of  this  equation  represents  the  family  of  ''inte- 
gral curves,"  as  seen  in  §  197 ;  in  many  cases  additional 
data  are  given  that  enable  us  to  determine  the  constant  of 
integration. 

Example :  Find  the  equation  of  the  curves  whose 
normal  always  passes  through  a  fixed  point. 

Let  us  take  the  fixed  point  as   origin  of  coordinates. 

The  slope  of  the  normal  at  (2;,  ^)  is j ;  but  since  the 

normal  passes  through  the  origin,  its  slope  is  ^-  Hence 
the  differential  equation  of  the  required  curves  is 

or 

xdx-[-  y  dy  =  0. 
Solving,  we  get 

x^  +  y^  =  c. 

The  only  curves  having  the  given  property  are  circles 
with  center  at  the  given  fixed  point. 

EXERCISES 

1.  Find  the  equation  of  the  curves  whose  subnormal  is  constant. 
Draw  the  figure.     (See  Ex.  22,  p.  32 ;  cf .  also  Ex.  8,  p.  31.) 

2.  Find  the  equation  of  the  curves  whose  subtangent  is  constant 
and  equal  to  a.     Draw  the  figure.  ^^^       _  ^^^^ 

3.  Determine  the  curves  in  which  the  normal  at  any  point  is 
perpendicular  to  the  radius  vector  (i.e.  the  line  joining  the  point 
to  the  origin). 

4.  Determine  tlie  curves  in  which  the  perpendicular  from  the 
origin  upon  the  tangent  is  equal  to  the  abscissa  of  the  point  of 
contact. 


DIFFERENTIAL  EQUATIONS  OF  FIRST  ORDER     303 

5.  Determine  the  curves  in  which  the  area  inclosed  between  the 
tangent  and  the  coordinate  axes  is,  equal  to  n'^. 

6.  Determine  the  curves  such  that  the  area  included  between  the 

curve,  the  coordinate  axes,  and  any  ordinate  is  proportional  to  the 

ordinate.  .  ' 

Ans.  y  =  ce". 

7.  Find  the  curve  of  Ex.  6  that  crosses  the  ^-axis  (a)  at  (0,  2  a)  ; 
(6)  at  an  angle  of  45°. 

MISCELLANEOUS   EXERCISES 
Solve  the  following  equations. 

1.  x^dy  —  (1  +  xhj)dx  =  0. 

2.  V  —  =  1  —  v'^.     Solve  in  two  ways. 

dx 

3.  dy  —  sin  x  dx  =  2y  dx. 

4.  y  dx  -f  dy  =  y'^  dx.     Solve  in  two  ways. 

5.  (x  —  y)dx  +  (1  —  X  —  2  y)dy  =  0. 

6.  dy  +  x^y  dx  =  0.     Solve  in  two  ways. 

7.  —  =  a  —  cos  kt. 
dt 

8.  ^  +  (log  V  —  \)dx  =  0.     Solve  in  tw^o  ways. 

y 

9.  y  =  1  -  y  +  sin  t.  10.    (a:2  -  4  xy)dx  +  y"^  dy  =  0. 
11.    x^  -y  =  xVx^  +  y\  12.    (1  +  x^)dy  -  (1  +  xy)dx  =  0. 

13.  'I^  =  a^-  kH\  14.    v  —  =y-v. 

dt  dy 


CHAPTER   XXVI 

DIFFERENTIAL  EQUATIONS  OF  HIGHER  ORDER 

I.   Introduction 

206.  General  and  particular  solutions.  Being  given  a 
relation  between  x^  y^  and  n  arbitrary  constants,  say 

(1)  ^(^.  y,  G^  •••,0=^' 

let  us  differentiate  this  relation  n  times  in  succession. 
The  equations  thus  obtained  form  with  the  original 
equation  a  set  of  n  +  1  equations  from  which  the  n  con- 
stants may  be  eliminated.  The  result  is  a  differential 
equation  of  the  n-\\\  order, 

(2)  ^(:r,  ^,  y,  ...,y">)  =  0. 

Conversely,  corresponding  to  a  differential  equation  of 
the  form  (2),  there  exists  in  general  a  relation  of  the  form 
(1)  which  satisfies  the  differential  equation.  Equation 
(1)  is  called  the  general  solution  of  equation  (2).  Thus 
the  general  solution  of  a  differential  equation  of  the  n-th  order 
involves  n  arbitrary  constants. 

It  is  understood  that  the  general  solution  contains  n 
essential  constants :  i.e.  that  it  cannot  be  replaced  by  an 
equally  general  form  containing  a  smaller  number  of  con- 
stants.    Thus  the  equation 

y  =  c^e^'^^^ 

appears  at  first  sight  to  contain  two  constants,  but  there 
is  really  only  one.     For,  writing  the  equation  in  the  form 

y  z=  c^e^  •  e*^2 
304 


DIFFERENTIAL  EQUATIONS  OF  HIGHER  ORDER     305 

and  setting 

c^e''^  =  (7, 

we  see  that  the  equation 

^  =  Ce"" 
is  equally  general. 

A  particular  solution  is  one  that  is  obtained  from  the 
general  solution  by  assigning  particular  values  to  one  or 
more  of  the  arbitrary  constants.  Thus  a  particular  solution 
may  contain  any  number  of  constants  less  than  the 
maximum  number,  n. 

For  example,  it  follows  from  Ex.  18,  p.  50,  that  the 
equation 

dt^ 
is  satisfied  by  the  equation 

x  =  A  cos  kt  -\-  B  sin  kt, 

where  A  and  B  are  arbitrary.  Since  the  differential 
equation  is  of  the  second  order,  the  solution  here  given, 
containing  two  constants,  is  the  general  solution.  Par- 
ticular solutions  are 

x=  A  cos  kU 

X  =  A(cos  kt  +  sin  A;f), 

x=^  sin  kt^ 

rc  =  0, 
etc. 

207.  Geometric  interpretation.  Given  a  differential 
equation  of  the  second  order,  and  of  the  first  degree  in  y"^ 

y"  =f(x,y,  y), 
we  may  in  general  assign  values  at  pleasure  to  x^  ?/,  and  y\ 
and  compute  the  corresponding  value  of  y".  The  equation 
is  satisfied  by  the  coordinates  of  any  point  (2:,  y')  moving 
in  any  direction,  provided  its  direction  is  changing  at  the 
proper  rate.  Or,  since  the  value  of  y",  together  with  the 
assumed  value  of  y\  determines  the  curvature  of  the  path, 
we  may  also  say  that  the  differential  equation  is  satisfied 


306  CALCULUS 

by  the  coordinates  of  any  point  moving  in  any  direction, 
provided  its  path  has  always  the  proper  curvature. 

The  paths  of  the  point  (re,  y)  moving  in  the  manner 
just  described  are  called,  as  in  §  197,  the  integral  curves 
of  the  given  differential  equation.  The  ordinary  equation 
of  the  family  of  integral  curves  is  of  course  the  general 
solution  of  the  differential  equation ;  since  this  solution 
contains  two  arbitrary  constants,  or  parameters,  it  follows 
that  the  integral  curves  form  a  doubly -iyijinite  system. 
The  point  (x^  ?/)  may  start  from  any  assumed  initial 
position  in  any  direction  ;  hence  through  any  point  in  the 
plane  there  pass  infinitely  many  integral  curves. 

The  above  discussion  is  readily  extended  to  differential 
equations  of  the  third  and  higher  orders. 


EXERCISES 

Find  the  differential  equation  whose  general  solution  is  as  follows. 

1.  y  =  c^  +  c^e'^='.  Ans.    y"  —  2  y'  =  0. 

2.  y  =  c\e^  +  c^e'^. 

3.  y  =  Cjfi^  +  c^xe^. 

4.  y  =  Ci  sin  x  +  c^  cos  x,  Ans.   y"  ■\-  y  —  0. 

5.  y  —  c\-\-  CgX  +  a;2. 

6.         ^     =     Cl(l      +      Xy^     +     Cg. 

Solve  the  following  differential  equations,  and  discuss  the  nature 
of  the  integral  curves. 

7.   y"  =  0.  8.   y"  =  1. 

9.    y"  =  6  ar.  10.    y"  =  y' . 

11.  ^^  +  -^'"3"  =  a.     (Cf .  Ex.  2,  p.  78.) 

2/  . 

12.  Solve  Ex.  8,  (a)  if  the  curve  touches  the  line  y  —2  x  2X  (1,2); 
(i)  if  the  curve  passes  through  the  points  (1,  2),  (3,  3)  ;  (c)  if  the 
curve  passes  through  (1,  1)  ;  {d)  if  the  curve  intersects  the  ly-axis  at 
right  angles.     Draw  the  curve  (or  several  of  the  curves)  in  each  case. 

13.  Solve  Ex.  9  for  each  of  the  cases  of  Ex.  12. 


DIFFERENTIAL  EQUATIONS  OF  HIGHER  ORDER     307 

II.     The  Linear  Equation  with  Constant 
Coefficients 

208.  The  linear  equation.  We  have  already  (§  203) 
defined  the  linear  equation  of  the  first  order  as  an  equa- 
tion that  is  of  the  first  degree  in  y  and  y^ .  More  gener- 
ally, a  differential  equation  of  the  n-t\\  order  is  said  to  be 
linear  if  it  is  of  the  first  degree  in  ?/,  ?/',  •••,  y^^^\  Thus 
every  linear  differential  equation  of  the  n-th  order  can  be 
written  in  the  form 

where  the  coefficients  /?i,  •"•,  Pn  ^^^^  the  right  member  X 
are  functions  of  x. 

In  what  follows,  we  shall  be  concerned  entirely  with  the 
important  special  case  in  which  the  functions  pi,  •••,  p„  are 
constants : 
(1)  y^^^  +  ay-^^  +  ...  +a,y  =  X. 

209.  The  homogeneous  linear  equation.  A  linear  differ- 
ential equation  whose  right-hand  member  is  0  is  said  to 
be  homogeneous.^  Thus  the  general  form  of  the  homo- 
geneous linear  equation  with  constant  coefficients  is 

(1)  ?/<'^)  +  ay-^^  +  -•  +a^y  =  0. 

This  equation  is  important  not  only  in  itself  but  because 
its  solution  must  be  determined  before  that  of  the  non- 
homogeneous  equation  (1)  of  §  208  can  be  found. 

If  1/ =  ?/j  is  a  particular  solution  of  equation  (1),  then 
y  =  ^1^1,  where  e^  is  arbitrary,  is  also  a  solution,  as  ap- 
pears at  once  by  substitution  in  (1).  Further,  if  y  =  y^ 
is  a  second  particular  solution,!  then   not    only  y  =  c^y,^ 

but  also 

y  =  e^y^  +  c^y^ 

*  That  is,  it  is  homogeneous  in  y  and  its  derivatives.     See  §  199. 
t  That  is,  a  solution  not  of  the  form  y  =  Ciyi. 


308  CALCULUS 

is  a  solution.     Finally,  if 

y  =  HVv 
y  =  ^2^2' 

y  =  CnVn 

are  n  distinct  particular  solutions,  then 

y  =  ^1^1 +  ^22/2+    •••    +^n«/n 

is  a  solution,  and  since  it  contains  n  arbitrary  constants, 
it  is  the  general  solution. 

We  proceed  to  show  that  the  general  solution  of  equa- 
tion (1)  can  always  be  written  down,  provided  a  certain 
algebraic  equation  of  the  n-\h  degree  can  be  solved.  The 
theory  will  be  developed  in  detail  only  for  the  equation 
of  the  second  order. 

210.  The  characteristic  equation.  The  homogeneous 
linear  equation  of  the  first  order,  viz., 

y'  +  HV  =  ^' 
is  evidently  satisfied  by 

y  =  e~^'^^ . 

This  suggests  the  possibility  of  determining  m  so  that 

y  _   gmx 

will  be  a  solution  of  the  equation 

(1)  y'^  +  a^y'  +  a^y  =  0. 
Substituting  in  (1)  the  values 

y  =  e"»^,  y^  zzzme^""^  y"  =  m^e'^^ 
and  bracketing  out  the  factor  e"*%  we  find  that  the  differ- 
ential equation  is  satisfied,  provided 

(2)  m?  H-  a^m  -\-  a^z=  0. 

Equation  (2)  is  called  the  characteristic  equatio7i*  cor- 
responding to  (1).     Thus 

y  z=  e'^^ 

is  a  solution  of  equation  (1)  if  and  07dy  if  m  is  a  root  of 
the  characteristic  equation. 

*  Also  called  the  auxiliary  equation. 


DIFFERENTIAL  EQUATIONS  OF  HIGHER  ORDER     309 

211.  Distinct  roots.  If  the  roots  w^,  m^  of  the  char- 
acteristic equation  are  distinct,  we  obtain  at  once  two 
distinct  particular  solutions  of  the  differential  equation, 

viz., 

y  =  e^i"",    y  =  e'^i''. 

Hence,  by  §  209,  the  general  solution  is 
(1)  y=  Cie»^^i*+  c^e^^"". 

Example:  Solve  the  differential  equation 

y"  -y'~'2y  =  (i. 
The  characteristic  equation  is 
m^  —  m  —  2=0, 
whence  w  =  2  or  —  1. 

Thus  the  general  solution  of  the  given  equation  is 

212.  Repeated  roots.  When  the  charactexistic  equation 
has  equal  roots,  the  method  of  the  previous  article  does 
not  give  the  general  solution.  For,  if  m-^  =  m^^  equation 
(1)  above  becomes 

y  =  {jj^"*!^  -f  (?2e"'i^ 

hence  the  solution  contains  only  a  single  constant,  and  is 
a  particular  solution. 

To  find  a  second  particular  solution,  let  us  try 

y  =  xe^^^^ 
whence  y^  =  e'^^^'i^m^x  +  1), 

yif  _  e'^^''(m^x  -\-  2  Wj). 
Substituting  in  the  differential  equation,  we  find  that 

y  =  xe^^^ 
will  be  a  solution,  provided 
(1)  {m^^  +  a^m^  +  a^yx  +  2  w^  +  «i  =  0. 

Now  the  coefficient  of  x  vanishes  because  m^  is  a  root  of 
the  characteristic  equation.       Further,  since  m-^  =  Wg,  it 


310  CALCULUS 

follows  that 


«! 

%=-^, 


or  2m^-\-  a^  =  0. 

Thus    (1)  holds,    and   y  —xe'^^'^   is    a    second    particular 

solution. 

Therefore  the  general  solution  of  the  differential  equa- 
tion is 

213.  Complex  roots.  If  the  characteristic  equation  has 
complex  roots  a  ±  z/9,  the  general  solution  takes  the  form 

(1)  =  e^^Qc^e^^'^  -\-  c^e-'^^^. 

Up  to  this  point  the  exponential  function  has  not  been 
defined  for  imaginary  values  of  the  exponent.  If,  however, 
we  expand  e''''  formally  in  Maclaurin's  series,  and  compare 
with  the  series  for  sin  x  and  cos  x^  we  obtain  the  relation 

(2)  e»^  =  cos  X  +  i  sin  x. 

In  the  theory  of  functions  of  a  complex  variable,  this 
formula  is  taken  as  the  definition  of  the  imaginary  expo- 
nential function. 

By  means  of  (2),  the  right  member  of  (1)  may  be  sim- 
plified.    For, 

gi^x  _  (3Qg  ^^  +  z  sin  ^x^ 

^-iPx  —  Qos  ^x  —  i  sin  fix. 
Whence  (1)  becomes 

y  =  g''^[((Tj  +  Cg)  cos  fix  +  z(^i  —  C2)  sin  fix^^ 
or,  if  we  place 

Cj  -f-  <?2  =  ^1  1  '^y^i  ~  ^2)  ~  ^2 
and  drop  the  accents, 

(3)  y  =  e^(Ci  cos  ^x  +  c^  sin  pjt) . 
Changing  again  the  meaning  of  c^  and  e^,  we  may  write 

(3)"  in  the  form 

y  =  c^e*^  cos  (^fix  -{-  ^2)? 
as  is  easily  verified.     This  form  is  to  be  preferred  in  cer- 
tain applications. 


DIFFERENTIAL  EQUATIONS  OF  HIGHER  ORDER     311 

EXERCISES 

Solve  the  following  differential  equations. 
1.   ^"_5y +  62/  =  0.  2.   y"  =  y. 

ax^      ax 
5.    //"  +  3^'  =  0.  6.    — =^•2a;. 

f/^2 

7.    ^  +  ^^  =  0.  8.   y"  +  ny'  +  2d>y  =  0. 

9.   y"  -  4?/'  +  4i/  =  0.  ^ns.    ?/  =  qe2x  ^  C2a;e2*. 

10.   ^.=  0.  .  11.   4^+4^'  +  r  =  0. 

12.    9/'  +  12/  +  4//  =  0.  13.   y"  -oy'^by. 

14.  y"  +  2ij'  +  by  —0.  Ans.   y  =  e~''{c^co&2  x  +  c^sm^x). 

15.  ?/"  -  4  ?/'  +  6  ?/  =  0.  16.    ^  =  -  k'^x. 

17.   y'  +  93/  =  0.  18.   y"  +  2y'  +  y  =  0. 

19.  8/'+  16/+  Qy  =  0. 

20.  Find  the  equation  of  a  curve  for  which  y"  =  y,  if  it  crosses 
the  3/-axis  at  right  angles  at  (0,  1). 

21.  Find  the  equation  of  a  curve  for  which  y"  =  —  y,  ii  it  touches 
the  line  ?/  =  x  +  1  at  (0,  1). 

22.  Determine  the  curves  for  which  the  rate  of  change  of  the  slope 
is  equal  to  the  slope. 

23.  In  Ex.  22,  find  the  curve  that  touches  the  line  y  =  2  a:  at  the 
origin.  Ans.  y  =  2e^  —  2. 

24.  In  Ex.  22,  find  the  curves  that  cross  the  y-3,xh  at  45°. 

25.  In  Ex.  22,  find  the  curve  that  passes  through  (0,  1)  and  ap- . 
proaches  the  negative  a:-axis  asymptotically. 

26.  Show  that  e '^  =  i,  e^'  =  -  1,  e-""'  =  1. 

27.  Derive  formula  (2)  of  §  213  by  comparison  of  the  Maclaurin 

series  for  e'^,  sin  x,  and  cos  x. 

28.  Show  that,  if  the  characteristic  equation  has  equal  roots  m^, 
the  equation 

/'  +  «i/y'  +  a^y  =  0 
can  be  reduced  to  the  form  z"  =  0  by  the  substitution  y  =  ze^^i",  and 
derive  the  result  of  §  212  from  this  fact. 


312  CALCULUS 

214.  Extension  to  equations  of  higher  order.  The  the- 
ory of  §§  210-213  is  readily  extended  to  equations  of 
higher  than  the  second  order.  We  give  the  results  with- 
out proof : 

Let  there  be  given  a  differential  equation 
(1)  ?/(«>  +  ay-^^  ^  ...  j^a„y=0. 

(a)  If  the  roots  Wj,  mg,  •••,  m^  of  the  characteristic 
equation 

m^ -\- a^m^'^ -\-  •••  +  a,^  =  0 

are  all  distinct,  the  general  solution  of  (1)  is 
y  =  c^e'"^^  +  c^e"^^-''  +   •  •  •  +  c^e"^n^. 
(6)   Corresponding  to  a  double  root  m-^^  the  terms  in  the 
general  solution  are 

y  =  c-^e^-^^  4-  c^xe^^^ ; 
corresponding  to  a  triple  root, 

y  =  c^e"H^  -|-  c^xe"^^^  -f-  c^x^e"^-^"^ ; 
etc. 

(c)  A  pair   of    complex  roots  a  ±  ^/3  give   rise    to   the 

terms 

y  =  e«^((?j  cos  ^x  +  ^2  sin  ^x} ; 

a  pair  of  double  roots  a  ±  iff  give  rise  to  the  terms 

y  =  e'^(^c^  cos  I3x  +  ^2  sin  fix  +  c^x  cos  fix  4-  o^x  sin  fix^ ; 

etc. 

EXERCISES 

Solve  the  following  equations. 
1.   y'"  _  7  y'  +  6  ?/  =  0.  2.    y"'  =  4  ?/. 

3.    ?/'"  =  y"  +  6  ;/.  4.   y^'^^  -  12  y"  +  27  y  =  0. 

A71S.  y  —  c^e^  +  c^e  ^  +  c^xe  *  +  c^x^e~''. 
6.    y"'  =  0.  7.    ?/(''^  -2y"  +  y  =  0. 

10.  ?/"  -  6  .?/"  +  13  ^'  =  0.     Ans.  y=c^^  e^''{c..2  cos  2x  +  c^  sin  2  a:). 

11.  '^  =  x.  12.    ^^  +  4^  =  0. 


DIFFERENTIAL  EQUATIONS  OF  HIGHER  ORDER     313 

13.  2/(*)  -  4  y'"  +  14  y"  -  2Q  y'  +  25  ^  =  0. 

14.  y'"  +  3  /'  +  3  ^'  +  ^  =:  0.         15.    y'"  -  2  y"  -  y'  +  2  ^  =:  0. 
16.    Prove  the  results  of  §  214  for  the  equation  of  the  third  order. 

215.   The  non-homogeneous  linear  equation.     Let  us  con- 
sider now  the  non-homogeneous  linear  equation 

(1)  y«)  +  ^y-i)+  ...  +a^y  =  X 

In  solving  this  equation,  the  first  step  is  to  write  down 
the  general  solution 

of  the  homogeneous  equation  obtained  from  (1)  by  making 
the  right  member  0.  The  quantity  Y  is  called  the  comple- 
mentary function. 

The  next  step  is  to  obtain,  by  any  means  whatever,  a 
particular  integral  of  (1), 

Then  the  equation 

y  =  Y  +  y 

is  a  solution  of  (1),  as  appears  at  once  by  substitution, 
and  since  it  contains  n  arbitrary  constants,  it  is  the  gen- 
eral solution. 

Various  methods  are  known  for  finding  the  particular 
solution  — 

y  =  y- 

The  method  given  below,  though  not  entirely  general,  is 
usually  the  best  method  when  it  applies,  and  it  is  avail- 
able in  nearly  all  cases  that  arise  in  the  simpler  appli- 
cations. 

We  begin  with  an 

Example  :  Solve  the  equation 

(2)  y"  —by'  -\-Qy  =  x  -\-  e^"^. 

The  complementary  function,  i.e.  the  solution  of    the 

IS 

Y  =  c^e^^  -\-  c^e^^. 


314  CALCULUS 

To  obtain  a  particular  integral  of  (2),  proceed  as  follows; 
Differentiating  twice,  we  obtain 

(3)  ?/^4)  _  ^yin  J^Qy"  =  4e2x^ 

Differentiating  again,  we  get 

(4)  ?/«>  -  5  ?/^4)  4_  6  y"f  =  8  e^^. 
Multiplying  equation  (3)  by  2  and  subtracting  from  (4), 
we  get  the  homogeneous  equation 

(5)  yib)  _  7  ya)  ^  16  y'"  _  12  7j"  =  0. 

It  is  easily  seen  that  the  complementary  function  Y 
forms  part  of  the  solution  of  this  equation  ;  hence  two  of 
the  roots  of  the  characteristic  equation 

m^  —  7  m*  +  16  m.^  —  12  w^  =  0 
are  m  =  2,  3.      The    other   roots  are  2,  0,  0,      Thus  the 
general  solution  of  (5)  is 

(6)  ^  =  c^e^""  4-  o^e^""  +  ^3  4-  o^x  +  e^  xe^"^. 

Let  us  substitute  y  in  the  original  equation  as  a  trial 
solution,  noting,  however,  that  the  terms  arising  from  the 
complementary  function  must  disappear  identically  after 
the  substitution,  so  that  it  is  sufficient  to  substitute* 

(7)  y  =  c^-{-  c^x  -{-  c^xe^"". 

We  have  ,  ,   o        2^  .       2t 

?/'  =  <?4  4-  z  Grxe^""  4-  ^5^  % 

y^^  =.  \  CrXe^^  4-  4  Cr^e^^. 
Substituting  in  (2),  we  find  that  (7)  will  be  a  particular 
integral  provided  the  equation 

4  erxe^""  4-  4  c^e^^  —  ^  c^  —  10  c^x^""    . 
—  5  c^e^^  4-  6  (?3  4-  6  c^  4-  6  CrX^""  =  x  +  e"^^ 
holds  identically  —  i.e.  for  all  values  of  x.     The  terms  in 
xe'^'^  destroy  each  other.     Equating  coefficients  of  the  other 
functions,  we  find  the  following  : 

Coefficients  of  e^^  :        4  c^  —  5  Cg  =  1. 
Coefficients  of  a: :  6  c^  =  1. 

Constant  terms  :       —  5  c^  4-  6  Cg  =  0. 

*  That  is,  we  place,  temporarily,  c\  =  C2  =  0. 


DIFFERENTIAL  EQUATIONS  OF  HIGHER  ORDER     315 
This  gives 

t?5  =  —  1,     ^4  =  Qt     <?3  =  ^Q' 

Substituting  in  (6),  we  get  as  the  general  solution  of  (2) 

Thus  the  method  consists  of  the  following  steps : 

(a)   Write  down  the  complementary  function. 

(6)  Differentiate  both  members  of  the  given  equation 
successively  until  the  right  member  becomes  0,  either 
directly  or  by  elimination.  The  original  equation  is  thus 
replaced  by  a  derived  homogeneous  equation  of  higher  order 
(equation  (5)  in  the  example). 

(c)  Write  down,  by  §  214,  the  general  solution  (equa- 
tion (6)  above)  of  this  derived  equation.  The  comple- 
mentary function  will  always  be  a  part  of  this  solution,  so 
that  certain  of  the  roots  of  the  characteristic  equation  are 
known  beforehand  ;  these  should  be  removed  at  once  by 
synthetic  division. 

(t?)  Of  the  arbitrary  constants  occurring  in  this  general 
solution,  those  belonging  to  the  complementary  function 
({?j,  c^  above)  will  remain  arbitrary  in  the  final  result  ; 
they  may  therefore  be  placed  temporarily  equal  to  0,  since 
we  are  trying  to  find  merely  a  particular  solution  of  the 
original  equation.  The  other  constants,  the  so-called 
superfluous  constants,  are  determined  by  substituting  the 
value  of  1/  in  the  original  equation  as  a  trial  solution  and 
equating  coefficients. 

It  is  clear  that  the  success  of  the  method  depends  on  our 
ability  to  reduce  the  right-hand  member  to  0  by  differen- 
tiation and  elimination,  as  in  the  above  example.  Hence 
JTand  its  successive  derivatives  must  contain  only  a  finite 
number  of  distinct  functions  of  x  The  method  therefore 
applies  whenever  X  contains  only  constants  or  terms  of  the 
form  2:%  e°%  sin"  ax,  cos"  ax,  or  products  of  these,  n  being 
a  positive  integer. 


316  CALCULUS 


EXERCISES 


1.  Check  the  result  of  the  above  example  by  differentiation. 
Solve  the  following  equations. 

2.  y"  -1  y'  +I2y  =  X.  Ans.    y  =  c^e^^  +  c^e'^''  +  ^^  ^  +  ih- 

Ans.    y  =  c^e3^  +  c^e^'^  +  ie^^. 


3. 

y"  -  5  y'  +  Q  y  =  e*^. 

4. 

y"  +  y  =  cos  2  X. 

5. 

?/"  -  5  ?/'  +  4  ?/  =  2  X  -  3. 

6. 

y"  •^y'={l+xy. 

7. 

d'^x 

—  =  cos  t  —  x> 

dt:^ 

8. 

y"  —  ^  !i'  +  ^y  =  cos  X  — 

9. 

^^^  +  4  X  :--_  sin  3  t  +  fi. 
df^ 

?^.      Ans.    y  =  c^e^^  +  c^e^  +  j^ocosa: 
—  x^o  sin  X  +  xe^"". 


A ns.   X  —  c■^^ cos  2  f  +  c^  sin  2t  —  I  sin  3 ^  + 1  i^  —  ^. 

10.  y"  —  2y'  +  y  =  xe"^.  Ans.    y  =  e^(c^  +  c^x  +  Ix^). 

11.  ?/"  +  //  =  1  +  2  cos^. 


12. 

dhj  , 

— -I  +  ?/  =  x  sin  x. 

yl«s.    y  =  Cj^C 

13. 

d^u 
=  u. 

dir" 

14. 

v=  0. 

dv^ 

15. 

y'"  _  ?jij"  +  2y  =  3a:-4 

16. 

/"-2/'  +  y  =e-. 

17. 

a; 

18. 

^ ns.  u  =  \v^  +  c-^  +  Co. 
J.n,s.    2/  =  ci  +  e^(c^  +  c^x  +  \  x^). 


19.  Prove  the  statement  that  the  complementary  function  corre- 
sponding to  the  original  equation  is  always  a  part  of  the  solution  of 
the  derived  homogeneous  equation. 

III.     Miscellaneous  Equations  of  the  Second 

Order 

216.  The  equation  y"  =/(a:).  In  this  section  we  con- 
sider various  classes  of  equations  of  the  second  order  which 
can  be  solved  by  special  devices. 


DIFFERENTIAL  EQUATIONS  OF  HIGHER  ORDER     317 

The  simplest  case  is  that  in  which  the  second  derivative 
is  a  function  of  the  independent  variable  : 

y"  =/(^)- 

This  equation  can  be  solved  directly  by  two  successive 
integrations.     In  fact,  it  is  obvious  that  the  equation 

can  be  solved  by  n  successive  integrations. 

217.    The  equation  y"  =f(y)-     An  equation  in  which  the 
second  derivative  is  a  function  of  the  dependent  variable, 

can  always  be  rendered  exact  by  introducing  the  integrat- 
ing factor  2  y'  dx  in  the  left  member,  and  its  equivalent 
2  d^  in  the  right  member : 

2y'y"dx  =  2f(y)dy. 
Integrating,  we  find 

y'^  =  '^JKy^dy  +  c,. 

After  extracting  the  square  root  of  both  sides,  we  have  a 
differential  equation   of  the   first  order,   and  of  the  first 
degree  in  y\  in  which  the  variables  can  be  separated. 
Example :     Solve  the  equation 

yZ 


Multiplying  through  by  2  y'  dx,  we  get 

Id 


2  y' y"  dx  =      ^ , 


whence 


y2  =  _    +  ^ 


1 


?/2 


'^c^y'^  —  1 


y  =  ± 

y 


318  CALCULUS 


Separating  variables,  we  have 

±  ^  c^y^  —  1 
whence 


^ — ^ —  dx^ 


±  '^c^y^  —  1  =  c^a:  4-  <?2' 

218.  Dependent  variable  absent.  An  equation  of  the 
second  order  in  which  the  dependent  variable  y  does  not 
occur  is  an  equation  of  the  first  order  in  y^  ;  it  may  there- 
fore be  solved  for  j/'  by  the  methods  of  Chapter  XXV. 
The  result  is  of  course  an  equation  of  the  first  order  in  y, 
which  in  turn  may  be  solved  for  y. 

The  problem  of  §  216  is  evidently  a  special  case  of  the 
present  one. 

Example  :     Solve  the  equation 

(1  +  x)y''  -  y  =  0. 


Setting 

y'  =  v,y  =^, 

we  have 

(1  +  x^dv  —  vdx  —  0, 

or 

dv           ^^      _  A 
V          1  -\-  X 

Hence 

log  V  —  log  (1  +  2-)  =  log  Cy, 

or 

v  =  c^(l  +^). 

Replacing  v  hj  y'^  and  integrating  again,  we  find 

c 
or,  with  (?j  in  place  of  -^, 

y  =  c^(\  +  xy  4-  ^2- 


DIFFERENTIAL  EQUATIONS  OF  HIGHER  ORDER     319 

219.  Independent  variable  absent.  An  equation  of  the 
second  order  from  which  the  independent  variable  x  is 
absent  may  be  written  as  an  equation  of  the  first  order  in 
the  variables  y  and  v  by  putting 

dy  d^y        dv 

dx  dx^        dy 

The  truth  of  this  last  formula  is  obvious  : 

d^y  _  dv  _  dv      dy  _     dv 
dx^      dx      dy      dx        dy 

It  should  be  noted  that  the  problem  of  §  217  is  merely 
a  special  case  of  the  present  one. 

Example  :     Solve  the  equation 


With 


this  becomes 


or 


Whence 


^"  =  i/y  • 


dy 


dv 
v—  =  yv, 
dy 


dv  =  y  dy^ 
V  =  ^y^  +  Cy 


^^l     =dx. 
and,  if  c^  >  0, 

^  /  —  arctan  — ^ —  =  a:  -f-  c 

C  2 

This  may  be  simplified  by  writing  -j-  in  place  of  c^ 
—  arctan  ^  =  a:  4-  ^2* 


320  CALCULUS 

EXERCISES 

Solve  the  following  equations. 


1. 

^'^  =  X8. 

dx^ 

3. 

..'"  =  «Y 

4. 

y"=  '• 

Vy 

dh          1 

5. 

dc^        fi 

7. 

x^y"  =  y'. 

9. 

y"  =  1  -  y'- 

11. 

y"  +  yy'  =  o. 

12. 

d-^y  ^  1  <^3/  _  0 

dx'^      X  dx 

d^x      -, 


Ans.  ax  =  log(y  +  Vy'^  +  c^)  +  Cg. 
6     — =:-! 

8.    /y^//"  =  7/. 
10.    ?/"  =  a:e*. 

Ans.  y  =  c^  log  X  +  c 


2' 
X  (IX 

13.    !^=  t-^^y. 
dt^  ^ 

U,    t—+^=l.  Ans.  X  =  t +  c,]ogt  +  c^. 

dt^      dt  1     &  - 

16.    y^ =  - .  Ans.   (x  -  ciY  +{y  -  c^y  =  a\ 

(1  +  y"^)l       « 

16.  ^  =  -  k^x. 
df^ 

17.  d^  =  _kH. 
dfi 

18.  Of   the   above   exercises,   which   ones   can   be   solved   by  the 
methods  of  section  II? 

19.  Solve  Exs.  16,  17  by  the  methods  of  section  II. 

20.  Show  how  to  solve  the  equation 

y"  +  Py'  +  Qy'^  =  0, 
where  P  and  Q  are  functions  of  x  alone. 


CHAPTER    XXVII 

APPLICATIONS   OF   DIFFERENTIAL   EQUATIONS   IN 

MECHANICS 

I.    Rectilinear  Motion 

220.  Rectilinear  motion.  Consider  a  point  P  moving  in 
a  straight  line :  for  instance,  the  centroid  of  a  falling 
body,  of  the  piston  of  a  steam  en-      q  p 

gine,    or  of   a  train   running    on  a       '*  x-         >'  ^ 

straight  track.    The  position  of  the  ^^^-  ^^^ 

point  at  any  instant  is  determined  by  its  abscissa  OP  =  x, 
counted  from  an  arbitrarily  chosen  origin  0  on  the  line,  a 
definite  sense  along  the  line  being  selected  as  positive. 

As  the  point  moves,  its  abscissa  2;  is  a  function  of  the 
time : 

X=(f>(t). 

If  this  function  is  known,  the  motion  of  the  point  is  com- 
pletely determined.     The  velocity  v  is  found  as  the  first 

dx 

derivative  — ,  and  the  acceleration  /  as  the  second  deriv- 

d  X 
ative  — — ,  of  the  abscissa  x  with  respect  to  the  time  (see 

§55). 

In  most  applications,  however,  it  is  the  converse  prob- 
lem that  presents  itself.  Thus,  the  velocity  may  be  given 
as  a  function  of  ^  or  a:  or  both,  say 

so  that  in  order  to  determine  the  position  of  the  point  at 
any  time  it  is  necessary  to  solve  this  differential  equation 
Y  321 


322  CALCULUS 

of  the  first  order.  Or,  and  this  is  the  most  common  case, 
the  acceleration  may  be  given  as  a  function  of  t^  x^  and  v 
(or  of  any  one  or  two  of  these),  say 

(1)  S^-^^*'"''"^- 

The  abscissa  x  is  found  in  terms  of  t  by  solving  this 
differential  equation  of  the  second  order. 

It  should  be  noted  that  when  the  acceleration  (or  the 
velocity)  is  given,  the  motion  is  not  completely  determined 
unless  "  initial  conditions "  are  also  given  by  means  of 
which  the  constants  of  integration  can  be  determined. 

221.  Motion  of  a  particle  under  given  forces.  Suppose 
the  ''  point "  whose  motion  was  discussed  in  the  preceding 
article  is  a  material  particle  moving  under  given  forces. 
If  the  particle  is  free  to  move  in  any  direction,  the  motion 
will  be  rectilinear  only  if  the  resultant  F  of  all  the  applied 
forces  lies  in  the  same  straight  line  with  the  initial 
velocity.  The  product  of  the  mass  by  the  acceleration  is 
equal  to  the  resultant  force,  by  §  187.  If  we  multiply 
both  members  of  equation  (1)  above  by  m,  and  write 
F(t^  x^  v)  in  place  of  m/(f,  x^  v},  that  equation  takes  the 
form 

drX 

m-^  =  F(t,x,v-). 

This  equation  and  equation  (1)  of  §  220  are  mathematically 
equivalent,  since  one  is  a  mere  constant  multiple  of  the 
other.  The  difference  lies  in  the  physical  meaning  of  the 
quantities  involved. 

It  should  be  noted  that  the  term  "  particle  "  as  here  used 
does  not  mean  necessarily  a  mere  mass-point.  The 
'•'•  particle  "  may  be  a  body  of  any  size  or  shape,  provided 
that  all  the  forces  acting  may  be  regarded  as  applied  at  a 
single  point,  and  that  the  motion  of  one  point  determines 
the  motion  of  the  whole  mass,  as  in  the  case  of  a  rigid 
body  moving  without  rotation. 


APPLICATIONS  OF  DIFFERENTIA^.  EQUATIONS     323 
222.    The  equation  of  motion.     The  equation 

(1)  mg  =  F(f,  X,  K), 

or  its  equivalent 

(2)  S=/(^^-'')' 

is  called  the  equation  of  motion.  It  follows  from  what  has 
been  said  that  the  rectilinear  motion  of  a  particle  is 
determined  by  the  equation  of  motion  together  with  the 
initial  conditions. 

In  each  problem  there  are  in  general  three  steps :  first, 
to  write  the  equation  of  motion  ;  second,  to  solve  this 
equation,  determining  the  constants  of  integration  in 
accordance  with  given  initial  conditions ;  third,  to  in- 
terpret the  results. 

When   the   forces    acting  are    given,    the    equation    of 

motion  can  be  written  at  once  :   we  have  only  to  equate 

ox 
w^— —  to  the  sum  of  the  components  of  the  forces  in  the  di- 

(X/b 

rection  of  motion. 

In  the  most  general  case,  the  equation  of  motion  may  be 
expressed  as  a  differential  equation  of  the  second  order  in 
X  and  t  by  substituting 

dx 
dt 

Special  cases,  however,  are   common.     If  the   force  is  a 

function  of  t  only,  the  method  of  §  216  evidently  applies. 

If  #  is  a  function  of  t  and  v,  we  may  use  the  method  of 

§  218,  writing 

^ON  d'^x  _  dv 

^  ^  'd^~~dt' 

If  jP  is  a  function  of  x  and  v,  the  method  of  §  219  applies : 
in  this  case,  since 

d^x  _  dv  _dv     dx 

dt^       dt      dx     dt 


324  CALCULUS 

we  substitute 

^  ^  dt^        dx 

We  shall  find  that  in  many  cases  a  variety  of  methods 

may  be  used. 

In  any  problem  we  may  desire  to  know  the  position  of 
the  particle  at  any  time,  the  velocity  at  any  time,  and  the 
velocity  at  any  position.  We  should  therefore  try  to 
obtain  three  equations,  giving  *  x  in  terms  of  <,  v  in  terms 
of  f,  and  V  in  terms  of  x^  respectively.  The  {x^  f)-equation 
is  of  course  obtained  by  solving  the  equation  of  motion 
(1)  (or  (2))  as  an  equation  in  x  and  ^,  and  determining 
the  constants.  The  (v,  ^) -equation  may  be  found  by 
differentiation  of  the  (a:,  Q -equation,  after  which  the 
(v,  a:) -equation  may  be  obtained  (theoretically  at  least) 
by  eliminating  t  between  the  other  two.  If  it  is  possible 
to  introduce  (3)  and  apply  the  method  of  §  218,  the 
(v,  ^) -equation  results  directly  from  the  first  integration  ; 
if  formula  (4)  and  §  219  can  be  used,  the  (v,  2:)-equation 
is  obtained  directly. 

223.  Uniformly  accelerated  motion.  A  motion  is  said  to 
be  uniformly  accelerated  if  the  applied  force,  and  hence  the 
acceleration,  is  constant  (cf.  §  ^b).  If  the  constant  ac- 
celeration be  denoted   by  h^  the    equation    of   motion   is 

simply 

di^x  _  7 

EXERCISES 

1.  Write  the  differential  equation  of  uniform  rectilinear  motion 
(§  55),  and  find  x  in  terms  of  t,  v  in  terms  of  t,  and  v  in  terms  of  x, 
if  a:  =  2  and  v  =  4  when  t  —  0.  Solve  the  equation  of  motion  in  three 
ways,  by  the  methods  of  §§  212,  216,  and  219,  and  obtain  the  (y,  t)- 
equation  and  the  {v,  a:)-equatiou  in  each  of  the  ways  suggested  in 
§  222.     Draw  the  graph  of  each  equation. 

*  Explicitly  if  possible. 


APPLICATIONS  OF  DIFFERENTIAL  EQUATIONS     325 

2.  Solve  Ex.  1  if  X  =  10  when  t  =  5  and  a;  =  22  when  t  =  9.  Find 
the  values  of  x  and  v  when  (  =  0. 

3.  The  velocity  of  a  particle  at  the  time  t  is 

u  =  6  ^  —  5. 

Find  (a)  the  acceleration ;  (b)  the  space  covered  in  4  seconds ;  (c)  the 
velocity  when  x  =  6  (x  being  measured  from  the  starting  point). 
Describe  the  motion  in  words. 

4.  The  velocity  of  a  particle  at  the  distance  x  from  the  starting 

point  is  

V  =Vx  +  10. 

Find  X  in  terms  of  /;  also  find  the  acceleration. 

5.  A  particle  falls  under  gravity,  all  resistances  being  neglected. 
Write  the  equation  of  motion,  taking  motion  downward  as  positive, 
and  solve  it  by  three  methods.  Explain  the  meaning  of  the  con- 
stants of  integration. 

6.  Determine  the  constants  of  integration  in  Ex.  6  if  the  particle 
falls  from  rest,  the  starting  point  being  taken  as  origin.  Draw  the 
graph  of  the  (x,  t)-  and  (v,  ^)-equations,  noting  that  the  latter  is  the 
first  derived  curve  of  the  former  (§  3.5). 

7.  (a)  Solve  Ex.  5  if  the  initial  velocity  is  10  ft.  per  second  up- 
ward, (b)  How  far  and  how  long  does  the  particle  rise  ?  (c)  Find 
V  and  t  when  the  particle  is  20  ft.  below  the  starting  point. 

Ai^s.    (c)  V  =  37.1  ft.  per  second. 

8.  Solve  Ex.  5  if  a:  =  10  when  t  =  1  and  x  =  100  when  t  =  3. 
Does  the  particle  at  first  move  upward  or  downward  ?  Find  the  ve- 
locity at  the  end  of  1  second.  Ans.    13  ft.  per  second. 

9.  If  a  stone  dropped  from  a  balloon  while  ascending  at  the  rate 
of  20  ft.  per  second  reaches  the  ground  in  10  seconds,  what  was  the 
height  of  the  balloon  when  the  stone  was  dropped  ?  With  what  ve- 
locity does  the  stone  strike  the  ground  ? 

10.  Solve  Ex.  5  if  the  velocity  2  ft.  below  the  starting  point  is 
23  ft.  per  second.  If  the  starting  point  is  500  ft.  above  the  earth's 
surface,  when  and  with  what  velocity  does  the  particle  reach  the 
earth  ?  Ans.   t  =  5  or  6i  seconds. 

11.  Show  that  the  velocity  acquired  by  a  body  falling  from  rest 
through  a  height  h  is  

V  =  V2  gh. 
Derive  the  formula  in  two  ways. 

12.  A  body  falls  50  ft.  in  the  third  second  of  its  motion.  Find  the 
initial  velocity. 


326  '     CALCULUS 

13.  A  body  falls  under  gravity.  Find  the  distance  covered  in 
6  seconds  if  at  the  end  of  2  seconds  the  distance  below  the  starting 
point  is  84  ft. 

14.  The  motion  of  a  railroad  train  is  uniformly  accelerated.  If 
when  the  train  is  250  ft.  from  a  station  the  velocity  is  30  ft.  per 
second,  when  600  ft.  from  the  station  it  is  40  ft.  per  second,  find  the 
acceleration,  and  the  velocity  when  passing  the  station. 

Ans.   Vq  =  20  ft.  per  second. 

15.  A  stone  is  thrown  vertically  upward  from  the  top  of  a  tower. 
At  the  end  of  2  seconds  it  is  400  ft.  above  the  ground,  and  is  still 
rising,  with  a  velocity  of  10  ft.  per  second.  Find  the  height  of  the 
tower.  Ans.   316  ft. 

16.  A  stone  thrown  upward  from  the  top  of  a  tower  with  a  velocity 
of  100  ft.  per  second  reaches  the  ground  with  a  velocity  of  140  ft.  per 
second.     Discuss  the  motion.     What  is  the  height  of  the  tower? 

Ans.  150  ft. 

224.  Momentum  ;  impulse.  When  a  particle  of  mass  m 
is  moving  with  a  velocity  v^  the  product  mv  of  the  mass 
by  the  velocity  is  called  the  momentum  of  the  particle. 

When  a  particle  moves  under  a  constant  force  F  from 
the  time  Iq  to  the  time  fj,  the  product  F(^t^  —  to)  of  the 
force  by  the  time  during  which  it  acts  is  called  the  impulse 
of  the  force  for  that  time-interval.  More  generally,  if 
F  varies  from  instant  to  instant,  let  us  divide  the  time 
from  ^Q  to  t^  into  n  equal  intervals  A^,  multiply  each  A^ 
by  the  value  of  F  at  the  beginning  (or  any  other  instant) 
of  the  interval,  and  form  the  sum  of  the  products  thus 
obtained.  The  limit  of  this  sum,  as  A^  approaches  0,  is 
the  impulse  of  the  variable  force  F  during  the  interval 
from  Iq  to  t-^ : 


/=  lim  y  FAt  =   f'Fdt. 


225.    The  principle  of  impulse  and  momentum.     Let  us 
write  the  equation 


APPLICATIONS  OF  DIFFERENTIAL  EQUATIONS  327 
in  the  form 

dV  XT 

dt 
Multiplying  by  dt  and  integrating  from  the  time  ^q,  when 
the  velocity  is  v^^  to  the  time  t^  when  the  velocity  is  v,  we 
find 
(1)  mv  —  tuVq  =1    Fdt 

By  §  224,  the  left  member  of  (1)  is  the  change  of 
momentum  in  the  time-interval  from  tf^  to  t^,  the  right 
member  is  the  impulse  of  the  force  F.  Henoe  we  have 
the 

Theorem  :  If  a  particle  moves  in  a  straight  line^  the 
change  of  momentum  in  any  time-interval  is  equal  to  the 
impulse  of  the  force  during  that  interval. 

This  theorem  will  be  referred  to  as  the  principle  of 
impulse  and  momentum. 

It  should  be  observed  that  what  we  have  really  done 
here  is  to  find  a  first  integral  of  the  equation  of  motion 
by  the  method  of  §  218.  Since  the  force  F  is  always 
either  directly  or  indirectly  a  function  of  ^,  the  above 
theorem  is  true  in  general ;  but  in  order  actually  to  com- 
pute the  impulse  directly  in  a  given  case,  the  force  must 
of  course  be  given  explicitly  as  a  function  of  t : 

F=F(t). 

If  the  force  F  is  constant,  equation  (1)  becomes  simply 
mv  —  'MVq  =  Ft—  FtQ. 

226.    Work.      When  a  particle  moves  in  a  straight  line 

under  the  action  of  a  constant  force  F,  the  work  done  is 

defined  as  the  product  of  the  force  by  the  distance  passed 

over  : 

W  =  Fx. 

When  the  force  is  variable,  we  proceed  as  follows: 
Take  the  line  of  motion  as  a:-axis,  and  suppose  the  body 
moves   from    x—  a  to  x=h.       Divide    the    interval   into 


'  i 


328  CALCULUS 

segments  Ax,  and  multiply  each  segment  A2:  by  the  value 
of  F  at  some  point  of  Ax.  The  limit  of  the  sum  of  the 
products  thus  obtained  is  defined  as  the  work  of  the  vari- 
able force  during  the  motion  : 


227.    The  principle  of  kinetic  energy  and  work.     Let  us 

write  the  equation 

d^x      TTT 
m  — -  =  J^ 

in  the  form 

dV  jy 

dx 

Multiplying  by  dx  and  integrating  between  the  a;-limits 
Xq  and  X  and  the  corresponding  v-limits  Vq  and  v,  we  find 

(1)  lmv^-lmv,^=  Crdx. 

By  §  135,  the  quantity  -|  mv^  is  the  kinetic  energy  of 
the  particle,  hence  the  left  member  of  (1)  is  the  change 
in  kinetic  energy  from  Xq  to  x.  By  §  226,  the  right 
member  is  the  work  done-  during  the  motion.  Hence  we 
have  the 

Theorem  :  If  a  'particle  moves  in  a  straight  line,  the 
change  of  kinetic  energy  in  any  space-interval  is  equal  to  the 
work  do7ie  hy  the  force  in  that  interval. 

This  is  the  principle  of  kinetic  energy  and  work. 

Here  we  have  merely  applied  to  the  equation  of  motion 

the  method    of    §  219.       In  order   to  compute  the  work 

directly,  the  force  must  of  course  be  given  explicitly  as  a 

function  of  a;  : 

F=F(x). 

If  the  force  is  constant,  equation  (1)  reduces  to 
J  mv^  —  }^  mv^  =  Fx  —  Fxq. 


APPLICATIONS  OF  DIFFERENTIAL  EQUATIONS     329 

EXERCISES 

1.  Verify  the  principle  of  impulse  and  momentum  in  Exs.  7,  8, 
p.  325. 

2.  Verify  the  principle  of  kinetic  energy  and  work  in  Exs.  7,  8, 
p.  325. 

3.  Solve  Ex.  14,  p.  326,  by  the  principle  of  kinetic  energy  and  work. 

4.  Solve  Ex.  15,  p.  326,  by  the  principles  of  §§  225,  227. 

5.  A  ball  of  mass  5^  oz.  strikes  a  bat  with  a  velocity  of  12^  ft. 
per  second,  and  returns  in  the  same  line  with  a  velocity  of  32  ft.  per 
second.  If  the  blow  lasts  -^^  second,  what  force  is  exerted  by  the 
batter?  Ans.    9  lbs. 

6.  A  ball  of  mass  5^  oz.  moving  at  50  ft.  per  second  is  caught  and 
brought  to  rest  in  a  distance  of  6  in.  AVhat  is  the  average  pressure 
on  the  hand?  Ans.   26  lbs. 

228.  Constrained  motion.  The  motion  of  a  body  some- 
times depends  on  other  conditions  than  the  given  forces. 
Thus,  the  piston  of  a  steam  engine  can  move  only  along 
the  cylinder,  a  body  sliding  down  an  inclined  plane  can- 
not fall  through  the  plane,  etc.  The  motion  in  such 
cases  is  said  to  be  constrained. 

In  the  case  of  constrained  motion,  let  the  applied  force 
be  resolved  into  components  along,  and  at  right  angles  to, 
the  path.  The  component  in  the  direction  of  motion  is  the 
"effective  force"  ;  the  motion  is  due  entirely  to  this  com- 
ponent, and  hence  it  is  only  this  component  that  appears  in 
the  equation  of  motion.  For  example,  when  a  particle 
slides  down  a  smooth  inclined  plane,  the  effective  force 
is  the  component  of  gravity  parallel  to  the  plane.* 

Further,  it  is  evident  that,  in  the  definitions  and 
theorems  of  §§  224-227,  the  force  F  must  be  taken  as 
merely  the  effective  component.  The  component  normal 
to  the  path  cannot  do  work,  or  contribute  to  a  change 
of  momentum. 

*  The  motion  is  supposed  to  take  place  alone;  a  ''line  of  greatest 
slope"  — i.e.  a  line  at  right  angles  to  a  horizontal  line  in  the  plane. 


330  CALCULUS 

EXERCISES 

1.  Write  the  equation  of  motion  down  an  inclined  plane,  and  solve 
it  in  a  variety  of  ways.     Explain  the  meaning  of  the  constants. 

2.  Determine  the  constants  in  Ex.  1  if  the  angle  of  inclination 
to  the  horizon  is  30°,  and  the  initial  velocity  is  (a)  0;  (b)  10  ft.  per 
second  up  the  plane.  In  (b),  how  far  and  how  long  will  the  body 
move  up  the  plane ?  A7is.    (b)  3}  ft. 

3.  A  bead  is  strung  on  a  smooth  straight  wire  inclined  at  45°  to 
the  horizontal.  What  initial  velocity  must  the  bead  be  given  to 
raise  it  to  a  vertical  height  of  10  ft.? 

4.  A  railroad  train  is  running  up  a  grade  of  1  in  200  at  the  rate 
of  20  miles  per  hour  when  the  coupling  of  the  last  car  breaks.  Fric- 
tion being  neglected,  (a)  how  far  will  the  car  have  gone  after  2  min- 
utes from  the  point  where  the  break  occurred  ?  (b)  When  will  it  be- 
gin moving  down  the  grade?  (c)  How  far  will  it  be  behind  the  train 
at  that  moment  ?  (c?)  If  the  grade  extends  1 500  ft.  below  the  point 
where  the  break  occurred,  with  what  velocity  will  it  arrive  at  the 
foot  of  the  grade?      Ajis.    (a)   2368  ft.;   (b)  3  minutes  3  seconds ; 

(c)  2689  ft. ;   (d)  25  miles  per  hour. 

6.  Show  that  it  takes  a  body  twice  as  long  to  slide  down  a  plane 
of  30°  inclination  as  it  would  take  to  fall  through  the  "  height "  of 
the  plane. 

6.  Show  that  in  sliding  down  a  smooth  inclined  plane  a  body  ac- 
quires the  same  velocity  as  in  falling  vertically  through  the  height 
of  the  plane. 

7.  A  mass  of  12  lbs.  rests  on  a  smooth  horizontal  table.  A  cord  at- 
tached to  this  mass  runs  over  a  pulley  on  the  edge  of  the  table ;  from 
the  cord  a  mass  of  4  lbs.  is  suspended.  Discuss  the  motion.  If  the 
12  lb.  mass  is  originally  5  ft.  from  the  edge  of  the  table,  find  when 
and  with  what  velocity  it  reaches  the  edge.  Check  by  the  principles 
of  §§  225,  227. 

8.  A  cord  hangs  over  a  vertical  pulley  and  carries  equal  weights 
of  10  lbs.  at  each  end.  If  a  1-lb.  weight  be  added  at  one  end,  discuss 
the  motion  of  the  system.     Find  v  when  the  system  has  moved  6  ft. 

229.  Simple  harmonic  motion.  If  a  point  P  moves  in 
a  circle  with  constant  angular  velocity  ©,  the  motion  of 
the  projection  P^  of  P  on  a  diameter  of  the  circle  is 
called  simple  harmonic  motion.     As  P  moves  in  the  circle 


APPLICATIONS  OF  DIFFERENTIAL  EQUATIONS     331 

uniformly,  P^  oscillates  from  A  through  0  to  B  and  back 

again.  p 

Suppose  Pj.  is  at  A  at  the  time  y^ 

f  =  0.     Then  in  time  t  the  angle  /            ^/ 

AOP   swept    out    by   the   radius       / ^ 

vector  of  P  is  equal  to  ojf,  hence  -^T             0     P^     lA 

the  distance  x  of  P^  from  0  is  V                        / 

(1)  x=  a  cos  (ot^  ^ -^ 

where  a  is  the  radius  of  the  circle.  ^^'  ^^^ 

If  when  t=  0  the  point  P  is  not  at  ^,  but  at  some 
point  P'  such  that  the  angle  AOP'  is  equal  to  e,  the 
equation  (1)  is  evidently  replaced  by 

(2)  x  =  a  cos  (^Q)t-\-  e'). 

The  abscissa  x  is  called  the  displacement  of  P^.; 
the  maximum  displacement  a  is  the  amplitude  of  the 
motion. 

The  time  of  completing  one  whole  oscillation  from  A 
to  B  and  back  is  called  the  period;  it  is  evidently  equal 
to  the  time  required  for  P  to  make  one  complete  revolu- 
tion, and  is  therefore 

CD 

The  number  of   oscillations  per  unit  time   is   called  the 
frequency ;   it  is  obviously  the  reciprocal  of  the  period  : 

n  =  -  =  -^ 

T      lir' 

The  angle  o)^  +  e  is  called  the  phase-angle^  or  simply  the 
phase,  of  the  motion. 

Differentiating  (2),  we  get  the  velocity 

dx 
v=  ——  —  aw  sin  (&)^  +  e), 

and  the  acceleration 

(3)  j  =  —-=-  aoi^  cos  ((ot  +  e). 


332  CALCULUS 

Combining  (2)  and  (3),  we  may  write  the  acceleration 
in  the  form 

— -  =  —  (o^x  : 

i.e.  the  acceleration  is  proportional  to  the  displacement.,  and 
is  always  directed  opposite  to  it. 

230.  Attraction  proportional  to  the  distance.  If  a  particle 
moves  in  a  straight  line  under  the  action  of  a  force  directed 
toward  a  fixed  point  0  in  the  line  of  motion,  and  propor- 
tional to  the  distance  x  from  that  point,  the  equation  of 
motion  can  evidently  be  written  in  the  form 

(1)  m  — —  =  —  mk'^x., 

where  A;  is  a  constant,  the  minus  sign  being  chosen  be- 
cause the  force  is  always  directed  opposite  to  the  displace- 
ment X.  The  fixed  point  toward  which  the  force  is 
directed  is  called  the  center  offeree. 

Integrating  equation  (1)  by  the  method  of  §  213,  we  get 

x=  c-^  cos  kt  H-  (?2  sin  kt^ 
whence 

Cm  'V 

V  =  —  =  —  ^(?isin  kt-{-  kccf  cos  kt. 
dt  ^  ^ 

Then 

2' 


Take  v  = 

0  and  X 

=  a  when  ^  =  0. 
a  =  (?^,  0  =  kc. 

whence 

c^  =  a,  (?2  =  ^» 

and  finally 

(2) 

x=  a  cos  kt, 

V  =  —  ak  sin  kt. 
Since  x  has  here  the  same  form  as  in  equation  (1)  of 
§  229,  it  follows  that  a  particle  moving  under  the  con- 
ditions of  this  article  performs  simple  harmonic  oscilla- 
tions about  the  center  of  force  0.  This  is  a  fact  of 
great  importance,  as  forces  directed  toward  a  fixed  point 
and  proportional  to  the  distance  from  that  point  are  of 
frequent  occurrence  in  nature. 


APPLICATIONS  OF  DIFFERENTIAL  EQUATIONS     333 

231.  Hooke's  law.  When  a  spiral  steel  spring  of  length 
AO=^l  is  stretched  to  a  length  AP  =  I  +  x,  the  tension 
in  the  spring,  or  the  force  tending  to  restore  it  to  its 
natural  length,  is  proportional  to  the  extension  x.  This 
law,  known  as  Hookes  law^  is  obeyed  very  closely  (pro- 
vided the  extension  is  not  too  great)  by  all  so-called  elastic 
materials. 

Suppose  a  steel  spring  of  negligible  mass  is  placed  on 
a  smooth  horizontal  table  with  one  end  fast  at  A.  Let 
the  natural  lens^th  of  the 

spring  be  A0=  ?.     A  par-  j^i I 9     ^     Qm 

tide   01   mass   m   attached 
to  the  free  end   is   drawn 

out  to  the  position  P  and  then  released.  The  only  force 
acting  is  the  tension  in  the  spring,  which  by  Hooke's  law 
is  directed  toward  the  position  of  equilibrium  0  and  is 
proportional  to  the  distance  from  0.  If  the  spring  offers 
the  same  resistance  to  compression  as  to  extension,  it  fol- 
lows from  §  230  that  the  particle  performs  simple  harmonic 
oscillations  about  0.     The  equation  of  motion  is 

Ct    X  7  9 

7n — -  =  —  mfc^x. 

Of  course  if  the  resistance  to  compression  is  not  the 
same  as  to  extension,  a  different  equation  comes  into  play 
as  soon  as  the  particle  passes  through  0. 

EXERCISES 

1.  In  the  problem  of  simple  harmonic  motion,  trace  the  curves 
showing  X,  v,  and  j  as  functions  of  t,  remembering  that  the  graph  of  v  is 
the  first  derived  curve,  the  graph  of  J  the  second  derived  curve,  of 
the  graph  of  x.     Take  rt  =  l,(o  =  2,  e  =  0. 

2.  Show  that,  if  x  performs  periodic  oscillations  as  in  §  229,  v  and 
/  do  likewise.  Prove  the  following  from  the  equations  of  §  229,  and 
verify  by  the  curves  of  Ex.  1 :  the  periods  of  all  three  are  the  same ; 
the  amplitude  of  v  is  o>  times  that  of  x,  the  amplitude  of  /  is  w  times 
that  of  y ;  in  phase,  v  differs  from  a:  by  ^  and  J  differs  from  v  by  ^  • 


334  CALCULUS 

3.  In  the  problem  of  §  230,  obtain  the  (y,  i)-equation  by  two 
methods.  Ans.     v  =  ±k  Va^  —  x^, 

4.  A  particle  has  simple  harmonic  motion.  Proceeding  from 
equation  (1)  of  §  230,  find  x  in  terms  of  <,  v  in  terms  of  i,  and  v  in 
terms  of  a:,  if  v  =  v^  and  a:  =  0  when  t  —  0. 

5.  Show  directly  from  equation  (2)  of  §  230  that  the  particle 
performs  periodic  vibrations  about  the  center,  and  find  the  amplitude 
and  the  period.  Find  when  and  where  the  velocity  is  a  maximum, 
and  find  the  magnitude  of  the  maximum  velocity. 

6.  A  steel  spring  offering  the  same  resistance  to  compression  as 
to  extension  is  placed  on  a  smooth  horizontal  table  with  one  end 
fixed.  The  spring  is  stretched  to  a  length  6  in.  greater  than  the 
natural  length  and  then  released.  Discuss  the  subsequent  motion  of 
a  mass  attached  to  the  free  end.     Take  Ic^  =  4.     Find  the  period. 

Ans.     T"  =  TT  seconds. 

7.  In  Ex.  6,  find  the  work  done  by  the  force  in  a  quarter-oscilla- 
tion.    Check  by  the  theorem  of  §  227. 

8.  Work  Ex.  6  if  the  steel  spring  is  replaced  by  a  rubber  band  of 
natural  length  1  ft.  Ans,     T  =  7.14  seconds. 

9.  In  Exs.  6  and  8,  discuss  the  effect  of  increasing  the  con- 
stant k^. 

10.    Work  Ex.  8  if  F  =  512.  Ans.     T  =  0.6  second. 

jij  11.    A  rubber  band  of  natural  length  AB  =  l  is  suspended 

vertically  with  a  weight  attached.  The  effect  of  the  weight 
is  to  stretch  the  band  to  a  length  AO  —  I  4-  h.  The  weight 
is  given  a  displacement  OP  =  a  and  then  released.  Write 
the  equation  of  motion  and  solve  it  completely.  Show  that 
the  particle  performs  simple  harmonic  oscillations  about  O, 
provided  a  <  A. 

12.  Solve  Ex.  11  \ia>h. 

13.  In  Ex.  11,  find  in  two  ways  the  work  done  by  the 
forces  as  the  particle  moves  from  P  to  0. 

14.  In  Ex.  11,  the  weight  is  let  fall  from  a  height  h 
above  B.  Determine  the  greatest  extension  of  the  rubber 
band. 

15.  A  bead  is  strung  on  a  smooth  straight  wire,  and  is 
'^"           attached  by  a  rubber  band  of  very  short  natural  length  to  a 

point  in  the  perpendicular  bisector  of  the  wire.     Taking  the  wire  as 
axis  of  2/,  show  that,  if  gravity  can  be  neglected,  the  equation  of 


/ 

J5-I- 

h 

0- 
a 


APPLICATIONS  OF  DIFFERENTIAL  EQUATIONS     335 

motion  of  the  bead  is  approximately 

Discuss  the  motion  completely. 

16.  A  particle  is  acted  upon  by  a  force  of  repulsion  from  a  point  O 
proportional  to  the  distance  from  0.  Neglecting  gravity,  write  the 
equation  of  motion  and  solve  it  completely,  taking  x  =  0  and  v  =Vq 
when  t  =  0.     Discuss  the  solution. 

17.  In  Ex.  16,  find  the  work  done  in  the  first  10  ft.  of  the  motion. 
Check  by  the  principle  of  kinetic  energy  and  work. 

18.  In  Ex.  16,  find  the  impulse  of  the  force  during  the  first 
second.     Check  by  the  principle  of  impulse  and  momentum. 

19.  It  is  shown  in  the  theory  of  attraction  that  the  attraction  of  a 
spherical  mass  on  a  particle  within  the  mass  is  directed  toward  the 
center  of  the  sphere  and  is  proportional  to  the  distance  from  the 
center.  Discuss  the  motion  of  a  particle  moving  in  a  straight  tube 
through  the  center  of  the  earth,  if  the  velocity  at  the  surface  is  0. 
Determine  the  proportionality  constant  k^^irom.  the  fact  that  the  force 
at  the  surface  is  —  mg. 

20.  In  Ex.  19,  how  long  does  it  take  the  particle  to  pass  through 
the  earth?  .4ws.  42^  luinutes. 

21.  A  straight  tube  is  bored  through  the  earth  connecting  two 

points  of  its  surface.     Show  that  the  equation  of  motion  of  a  particle 

sliding  in  this  tube  is 

d^x  mq 

m  —  =  -  -^2.  X, 

df^  R 

where  R  is  the  radius  of  the  earth  and  x  is  the  distance  of  the  particle 
from  the  midpoint  of  the  tube.  Discuss  the  motion.  Show  that 
the  time  of  passing  through  such  a  tube  is  independent  of  the  posi- 
tion of  the  endpoints. 

II.    Plane  Curvilinear  Motion 

232.  Rotation.  In  discussing  circular  motion,  it  is 
usually  convenient  to  take  as  dependent  variable  the 
angle  6  swept  out  in  the  time  t. 

The  problem  of  uniformly  accelerated  circular  motion 
(§  58)  is  closely  analogous  to  that  of  uniformly  accelerated 


336 


CALCULUS 


rectilinear  motion.     The  equation  of  motion  is  evidently 

where  k  is  the  constant  angular  acceleration. 

233.    The  simple  pendulum.     A  simple  pendulum  is  a 
point  swinging  in  a  vertical  circle  under  the  acceleration 

of  gravity. 

Let  P  be  a  particle  of  mass  m 
connected  to  the  point  0  by  a  cord 
or  rod  of  length  ?,  and  denote  by 
6  the  angle  between  OP  and  the 
vertical,  by  s  the  length  of  the  arc 
AP.  The  effective  force  acting  on 
P  is  the  component  of  gravity  tan- 
gent to  the  circle ;  since  this  is 
directed  opposite  to  s,  it  must  be  given  the  minus  sign. 
The  equation  of  motion  of  P  is  therefore 

ci)  '^'' 


Fig.  105 


m  — -  =  —  mg  sin  6, 

dv" 


But 


s  =  ie, 

so  that  (1)  may  be  written 
(2) 


ml — -  =  —  mg  sm  v. 


A  first  integration  of  (2)  can  be  performed  by  the 
method  of  §  219 ;  the  general  solution,  however,  cannot 
be  expressed  in  terms  of  elementary  functions.  We  shall 
therefore  consider  only  the  case  in  which  the  oscillations 
are  so  small  that  sin  6  may  be  replaced  by  0  (see  §  156), 
and  (2)  written  in  the  form 


(3) 


ml —  =  —  mqO. 
dfi  ^ 


This  equation  shows  that  for  small  oscillations  the 
motion  is  approximately  simple  harmonic.  The  remainder 
of  the  discussion  is  left  to  the  student. 


APPLICATIONS  OF  DIFFERENTIAL  EQUATIONS     337 

EXERCISES 

1.  Write  the  equation  of  uniform  circular  motion,  and  solve  it  in  a 
variety  of  ways,  explaining  the  meaning  of  the  constants.  Exhibit 
the  results  graphically. 

2.  Proceed  as  in  Ex.  1  for  uniformly  accelerated  circular  motion. 

3.  A  wheel  is  making  400  R.P.  M.  when  a  resistance  begins 
to  retard  its  motion  at  the  rate  of  10  radians  per  second.  When 
will  it  come  to  rest?  How  many  revolutions  will  it  make  before 
stopping  ? 

4.  Solve  equation  (3),  §  233,  taking  0  =  Oq  and  v  =  0  when  t  =  0. 
For  convenience,  put  ^  =  k'\ 

6.  Show  that,  in  the  problem  of  the  simple  pendulum,  the  time  of 
one  swing  or  beat  is 

9 

6.  Find  the  length  of  the  "seconds  pendulum"  —  i.e.  a  pendulum 
making  one  swing  per  second  —  at  a  place  where  g  =  32.17. 

Ans.  3.2595  ft. 

7.  Find  the  angular  velocity  o  in  terms  of  6  if  the  oscillations  are 
so  large  that  (2),  §  233,  must  be  used. 

8.  Study  the  motion  of  a  pendulum  making  small  oscillations,  if 
the  resistance  of  the  air  is  proportional  to  the  velocity. 

234.  The  equations  of  motion.  In  the  general  case  of 
motion  in  a  plane  curve,  it  is  convenient  to  resolve  all 
the  applied  forces  into  components  parallel  to  the  coord  i- 

nate  axes.     The  product  m  — -  of  the  mass  by  the  a^com- 

ponent  of  the  acceleration  (see  §  59)  is  equal  to  the  sum 
Fx  of  the  2;-components  of  all  the  forces;  similarly  for  the 
y-components.  We  thus  have  the  two  equations  of 
motion : 


■F:,, 

-F„. 

338  CALCULUS 

In  the  most  general  case,  both  F^  and  Fy  are  functions 

of  x^  y^  t,  and  the  velocity-components  i;^  =  — ,  ?;  = -^. 

dt  dt 

We  shall,  however,  confine  our  attention  to  the  case  in  which 
F^  is  a  function  only  of  x^  v^.,  and  ^,  and  Fy  is  a  function 
of  7/,  Vy^  and  t.  In  this  case  the  two  equations  of  motion 
may  be  integrated  separately.  We  thus  obtain  two  equa- 
tions giving  respectively  x  and  y  in  terms  of  t ;  these  are 
parametric  equations  of  the  path  of  the  moving  point. 
By  the  same  methods  as  those  already  used  we  find  equa- 
tions giving  v^  and  Vy  in  terms  of  ^,  and  in  terms  of  x  and 
y  respectively.  The  total  velocity  v  may  be  found  by 
§  57. 

235.  Projectiles.  A  simple  example  of  curvilinear 
motion  is  furnished  by  a  projectile  moving  under  gravity 
alone  —  i.e.  in  a  medium  whose  resistance  can  be 
neglected. 

Let  a  particle  be  projected  with  an  initial  velocity  v^ 
inclined  at  an  angle  a  to  the  horizontal.  With  the  start- 
ing point  as  origin  and  the  ^-axis  positive  upward,  the 
initial  conditions  are 

2:  =  0,  ?/  =  0,  Vx  =  VqCOs  a,  Vy  =  Vq  sin  a  when  f  =  0. 

The  force  of  gravity  acts  vertically  downward ;  there  is 
no  horizontal  force.     Hence  the  equations  of  motion  are 

d^x      f.  d^y 

These  may  be  integrated  and  the  constants  determined 
precisely  as  in  our  earlier  work. 

EXERCISES 

1.  Solve  the  problem  of  §235  completely,  finding  a:,  y,  v^,  and  Vy 
in  terms  of  t,  v^  in  terms  of  x,  and  Vy  in  terms  of  y. 

2.  In  Ex.  1,  by  eliminating  t  from  the  (x,  t)-  and  (y,  ^) -equations, 
show  that  the  path  is  a  parabola. 


APPLICATIONS  OF   DIFFERENTIAL  EQUATIONS     339 

3.  Show  that,  in  the  ideal  case  of  §  235,  where  all  resistances  are 
negligible,  a  projectile  whose  initial  velocity  is  horizontal  will  strike 
the  ground  in  the  same  time  as  a  body  let  fall  from  rest  from  the 
same  height. 

4.  The  range  of  a  projectile  is   the  distance   from   the  starting 

point  to  the  point  where  it  strikes  the  ground.     Show  that  the  range 

on  a  horizontal  plane  is  2 

i2  =  ^  sin  2  a. 
9 

5.  What  elevation  gives  the  greatest  range  on  a  horizontal  plane  ? 

6.  The  time  of  flight  is  the  time  from  the  starting  point  until  the 
projectile  strikes  the  ground.  Show  that  on  a  horizontal  plane  the 
time  of  flight  is 

9 

7.  A  stone  is  thrown  horizontally  from  the  top  of  a  tower  400  ft. 

high,  with   a  velocity  of  20  ft.  per  second.      (a)  AVhen,  (h)  where, 
and  (c)  with  what  velocity  does  it  strike  the  ground  ? 
Ans.    (a)  5  seconds  ;  (c)  161.2  ft.  per  second,  at  7°  8'  to  the  vertical. 

8.  Find  the  work  done  by  gravity  in  Ex.  7. 

9.  A  stone  slides  down  a  roof  sloping  30°  to  the  horizon,  through 
a  distance  of  12  ft.  If  the  lower  edge  of  the  roof  is  50  ft.  high, 
(a)  when,  (h)  w^here,  (c)  with  what  velocity  does  the  stone  strike 
the  ground?  Ans.  (6)  25.1  ft. 
from  the  building ;  (c)  59.7  ft.  per  second,  at  16"  30'  to  the  vertical. 

10.  A  pitcher  throws  a  ball  with  a  speed  of  100  ft.  per  second,  the 
ball  leaving  his  hand  horizontally  at  a  height  of  5  ft.  Show  that 
under  the  assumptions  of  §  235  the  ball  would  strike  the  ground 
before  reaching  the  batter  60  ft.  away. 

11.  A  particle  slides  on  a  smooth  roof  inclined  at  45°  to  the  hori- 
zontal. If  the  initial  velocity  is  10  ft.  per  second  parallel  to  the  edge 
of  the  roof  and  the  starting  point  is  20  ft.  above  the  edge,  find  when, 
where,  and  with  what  velocity  the  particle  leaves  the  roof. 

12.  A  particle  moves  under  the  action  of  a  force  directed  toward 
the  origin  0  and  proportional  to  the  distance  from  0  (cf.  §230). 
If  the  initial  conditions  are  a:  =  10,  y  =  0,  v^  =  0,  Vy  =  20  ft.  per 
second,  discuss  the  motion  completely.     Take  k  =  1. 

13.  Find  the  cartesian  equation  of  the  path  in  Ex.  12. 

14.  In  Ex.  12,  find  the  work  done  in  one  quarter  of  the  period. 
Check  by  the  principle  of  kinetic  energy  and  work. 


INDEX 


(The  references  are  to  pages.) 


Acceleration,  80,  279,  322,  323 

angular  — ,  84,  335 

components  of  — ,  86-88 

in  curvilinear  motion,  85,  337 

of  gravity,  81 
Algebraic  functions,  3 

differentiation  of  — ,  19-25 
Arc 

centroid  of  — ,  188 

of  a  plane  curve,  167 

of  a  space  curve,  250 
Area  (plane),  121,  148,  175;  176 

centroid  of  — ,  185 

in  cartesian  coordinates,  151 

in  polar  coordinates,  154 
Asymptotes,  97 

tests  for  — ,  98,  99,  105 

Catenary,  107,  153,  160,  168,  170 
Center  of  mass,  181 
Centroid,  180,  276 

of  arcs,  188 

of  areas,  185 

of  surfaces,  189 

of  volumes,  187 
Characteristic  equation,  210 
Composition  of  ordinates,  106 
Concavity,  33-36 
Conjugate  point,  95 
Constant  of  integration,  118,  143,  287 

determination  of  — ,  117-118,  322 
Continuity,  10,  202 

of  a  function  of  two  variables,  237 
Critical  point,  33 
Critical  value,  33 
Curvature,  76,  306 

center  of  — ,  78 

circle  of  — ,  78 

radius  of  — ,  78 
Curve  tracing,  32,  101,  105 

by  composition  of  ordinates,  106 

in  polar  coordinates,  114 
Cusp,  95 
Cycloid,  108,  111,  154,  161,  168,  170 


Definite  integral,  143 

as  limit  of  a  sum,  150 

change  of  variable  in  — ,  145 

fundamental  theorem  for  — ,  150 

geometric  meaning  of  — ,  144 
Density,  179,  274 

linear  — ,  179,  275 

surface  — ,  180,  274 
Derivative,  14 

as  quotient  of  differentials,  70 

geometric  meaning  of  — ,  15 

higher  — ,  18,  73 

of  a  function  of  a  function,  21 
Derived  curves,  42,  81,  325 
Difference-quotient,  15 
Differential,  70 

exact  — ,  296 

of  arc,  76 

of  independent  variable,  70 

total  — ,  240 
Differential  equation,  286 

exact  — ,  296 

general  solution  of  — ,  287,  305 

geometric      meaning     of  — ,     290, 
304 

linear  — ,  298,  300,  307 

order  of  — ,  287 

partial  — ,  286 

particular  solution  of  — ,  289,  305 

singular  solution  of  — ,  289 
Differentiation,  14 

of  algebraic  functions,  19-25 

of  exponential  functions,  62 

of  implicit  functions,  26,  73,  241- 
243 

of  inverse  trigonometric  functions, 
53 

of  logarithms,  58 

of  trigonometric  functions,  46-49 

standard  formulas  of  — ,  65-66 
Double  integrals,  260,  262,  264 

in  polar  coordinates,  265 

transformation  of  — ,  266 
Double  point,  95 


341 


342 


INDEX 


Envelope,  252 

of  normals,  256 

of  tangents,  254 
Epicycloid,  109 
Equation  of  motion,  323 

in  curvilinear  motion,  337 
Evolute,  256 
Exponential  function,  56 

differentiation  of  — ,  62 

graph  of  — ,  56 

imaginary  — ,  310 

Force,  279,  322,  323,  327 

distributed  — ,  280 

parallel  — ,  283 
Function,  1 

anti-hyperbolic  — ,  65 

branches  of  — ,  4,  51 

continuous  — ,  10,  19,  32,  121,  237 

differentiate  — ,  15,  19,  32 

homogeneous  — ,  294 

hyperbolic  — ,  64 

impUcit  — ,  26,  73,  92,  241-243 

increasing  and  decreasing  — ,  32 

integrable  — ,  121,  151 

inverse  — ,  27,  51,  56 

kinds  of  — ,  3 

limit  of  — ,  8,  11,  12,  237 

of  several  variables,  236 

one-valued  — ,  3,  19,  32 

rate  of  change  of  — ,  5,  15 

(See  also  algebraic  — ,  exponential 
— ,  etc.) 

Geometric  addition,  82 
Geometric  derivative,  85 
Graphic  solution  of  equations,  107 

Heterogeneous  masses,  273 
Hooke's  law,  333 
Hypocycloid,  110 

of  four  cusps,  111,  160,  168,  170, 
173 

Implicit  functions,  26,  92 

differentiation  of  — ,  26,  73,  241- 
243 

of  several  variables,  242-243 
Improper  integrals,  175-177 

geometric  meaning  of  — ,  177 
Impulse,  326,  327 
Indefinite  integral,  116-118 

geometric  meaning  of  — ,  121 
Indeterminate  forms,  202-205 


Infinitesimals,  8,  240 
a  theorem  on  — ,  159 
limit  of  ratio  of  — ,  8,  15,  70 
order  of  — ,  68 
principal  — ,  8 
principal  part  of  — ,  68,  70,  240 

Infinity,  11,  12 

Integral  curves,  291,  306 

Integral  tables,  172 

Integrating  factor,  297,  299,  317 

Integration,  116 
by  parts,  132 
by  substitution,  123,  134 
change  of  variable  in  — ,  122,  145 
of  rational  fractions,  137-141 
standard  formulas  of  — ,  126-127 

Inverse  trigonometric  functions,  51 
differentiation  of  — ,  53 
graphs  of  — ,  51-52 

Isolated  point,  95 

Kinetic  energy,  198,  328 
of  a  rotating  body,  198 

Law  of  the  mean,  201-202 
Limit,  6 

evaluation  of  — ,  202-208 

of  function,  8,  11,  12 

of  function  of  two  variables,  237 

of  ratio  of  infinitesimals,  8 

of  sin  a/a,  47 

the  —  e,  60,  230 

theorems  on  — ,  7,  159 
Line   integrals,    163,    167,    168.    170, 
188,  189 

evaluation  of  — ,  165-167 

fundamental  theorem  for  — ,  165 

geometric  meaning  of  — ,  164 
Logarithms,  56 

common,  63 

computation  of  — ,  234 

differentiation  of  — ,  58 

graph  of  — ,  56 

Napierian  — ,  60 

natural  — ,  60 

properties  of  — ,  57-58 

Maxima  and  minima,  33,  102 

applications  of  — ,  37-42 

in  polar  coordinates,  114 

tests  for  — ,  33-36 
Moment 

of  force,  283-284 

of  inertia,  190-198,  277 

of  mass,  180,  276 


INDEX 


343 


Momentum,  326,  327 
Motion 

circular  — ,  83,  335 

constrained  — ,  329 

curvHinear  — ,  82,  85-88,  335-339 

equation  of  — ,  323,  337 

plane  — ,  80-88,  321-339 

rectilinear  — ,  80-81,  321-335 

simple  harmonic  — ,  330-333,  336 

uniform  — ,  55,  324 

uniformly  accelerated  — ,  56,  324 

Node,  95 
Normal,  29 

length  of  — ,  30,  32 

to  a  surface,  245 
Normal  plane,  249 

Osculating  circle,  78 

Pappus,  propositions  of,  186,  189 
Parametric  equations,  72,  83,  167,  253 
Partial  derivatives,  237 

geometric  meaning  of  — ,  238 

higher  — ,  238 
Partial  fractions,  137 
Pendulum,  simple,  336 
Point  of  inflection,  35,  36,  102 

tangent  at  — ,  35 
Point  of  osculation,  95 
Pressure,  280-281,  284 

center  of  — ,  284 

fluid  — ,  281 
Projectiles,  338 

Radius  of  gyration,  190,  191 

Radius  of  inertia,  190 

Rate  of  change,  5,  15,  33,  80,  84,  88 

of  derivative,  33 
Rectification  of  curves,  167,  250 
RoUe's  theorem,  200,  226 
Rotation,  83,  335 

Series,  209 

absolute  convergence  of  — ,  219 
alternating  — ,  218,  229 
computation  by  — ,  228,  234 
convergence  and  divergence  of  — , 

211,  220 
geometric  — ,  6,  209,  210,  211 
Maclaurin's  — ,  222 
of  n  terms,  209 
power  — ,  220 


sum  of  — ,  210 

Taylor's  — ,  223 

tests  for  convergence  of  — ,  211- 
218,  220 

transformation  of  — ,  230-232 
Singular  points,  92 

kinds  of  — ,  95 
Slope  of  curve,  5,  15,  32 

polar—,  112 
Subnormal,  30,  32 
Subtangent,  30,  32 
Surfaces 

angle  between  lines  and  — ,  240 

angle  between  — ,  246 

centroids  of  — ,  189 

cylindrical  — ,  170 

general  — ,  268 

normal  to  — ,  245 

of  revolution,  168 

tangent  plane  to  — ,  244 
Symmetry,  101,  114 

Tangent,  29 

determination  of  —  by  inspection, 
93 

inflectional  — ,  35,  102 

length  of  — ,  30,  32 

stationary  — ,  35 

to  space  curve,  248,  250 
Tangent  plane,  244 
Taylor's  theorem,  227 
Time-rates,  88-91 
Trigonometric  functions,  45 

differentiation  of  — ,  46-49 

elementary  properties  of  — ,  45 

graphs  of  — ,  45-46 
Triple  integrals,  270-272 
Triple  point,  95 

Vector,  82 

components  of  — ,  82 

derivative,  85 

resultant  of  — ,  82 
Velocity,  80,  84,  320,  324 

angular  — ,  83,  84 

components  of  — ,  83 

in  curvilinear  motion,  82,  338 
Volumes 

centroids  of  — ,  187,  276 

general  — ,  161,  258-262,  265,  272 

of  revolution,  156-158,  266 

under  a  surface,  258-262,  265 

Work,  327,  328 


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